Monthly Archives: September 2014

UNYPAD Opens Office in Zamboanga

ZAMBOANGA City (September 29, 2014) – The Zamboanga City Chapter of the United Youth for Peace and Development (UNYPAD) under the leadership of Zulfhicar Samsaraji, UNYPAD Regional Coordinator for Western Mindanao,  established office at Suterville, Zamboanga City. Samsaraji believed that the said establishment of office in Zamboanga will make organization more visible and help implement

A meeting with IMAN


Last Sunday evening I had another wonderful time meeting familiar faces in this rather new world I am now affiliated with: The world of Medicine. And what’s better? I was able to become a part of a growing ‘family’ of professionals with a two unifying ideologies: Medicine and Islam.

The Islamic Medical Association of the Philippines (IMAN Phil) is a non-profit organization founded in 2007 with the aim to create a massive network of Muslim Professionals and Specialists in the medical field, uniting them not just because they have a single faith (Islam) and profession (doctors), but with the single ideology of serving the ummah of this generation. You can check the website of IMAN here:

The dinner meeting I attended was nothing but a simple informal gathering, organized by IMAN to meet-and-greet the Muslim Medical students in the Manila area. Although only a few of us were able to attend (most of the invited were on duty), it was still a wonderful gathering. The members of the boards of IMAN present who organized this meeting were Dr. Naheeda Mustofa (Internal Medicine, Clinical Nutrition; St. Lukes) and Dr. Khasmin Ismael (Board and founder, Health for Mindanao). Both Muslimah doctors inspired us with their experiences in the medical field from being a Muslim medstudent to being doctors. Also present in the meeting was my dear uncle, Dr. Nelson Laja (Medical Oncology, UP PGH) who also shared great insights about how we can further improve our

Arabic Kitchen has this nice place where we can gather around, without chair or high tables. 🙂 Arabian style
 [photo from Dr. Mustofa]

The ever outnumbered male doctors (and soon to be)
 [photo from Dr. Mustofa]
We also talked about the Visions, Missions and goals of IMAN, the past activities they had, the challenges they experienced, and also the plans and activities the IMAN plans to have in the  future in sha Allah. IMAN emphasized the importance of involving the student body to better empower the future generation of doctors. They also assured the students that they can find help and support from IMAN if ever they need to. In fact, a number of IMAN’s future plans were directed in helping the medstudents like providing medical books they need during their academic career. We are hoping to strengthen the network of Muslim medical students not just in the capital region but in the whole Philippines as well in the future, in sha Allah.

We, the students and interns who attended the meeting were all enthusiastic and excited with the future activities IMAN have and in sha Allah hopefully we could participate actively in all of those planned activities. Looking forward to meeting new faces and members of this growing family I am happily affiliated with. Perhaps my dream of attending an authentic “Muslim Medical Students Summit” is not a farfetched dream after all 😀

May Allah make things easy for IMAN. Allahumma ameen.

You can check the IMAN website here: or visit the IMAN FB Page: #no link yet#

For more info you can contact IMAN Phil at

Salam Kasilasa!
-Anakiluh, MD

Liham Kaibigan

Ikaw,At ngayon hindi ka na naman mahagilap, sa cellphone, sa email, maging sa sarili mong website. Ganito rin ang kuwento mo sa akin  noon, na isinasara mo ang sarili mo sa daigdig kung mayroong isang karanasan na bumagabag sa iyo. Ito marahil ang…

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The Tam Tam Project

Sometime in early March, I took in a stray dog. His name is Tam Tam. For nights at a time, I saw him sleeping on the pavement lining the front of the stores right outside my mother’s house. He was scrawny, ears riddled with dried pus. He had a putrid smell and lost almost all his fur to demodectic mange. Two-three nights in a row, I brought him food, which he ate like there’s no tomorrow, until I couldn’t take it anymore and took him in. This is Tam Tam on day one. This is him getti […]

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norte is philippine bet to oscars

The Film Academy of the Philippines has chosen Lav Diaz’s opus Norte: End of History to represent the country in the bid for a nomination for best foreign-language film at the Oscars next year.
Ahead of the October 1 submission deadline for countries to name their official submission, 2014’s field of international Oscar hopefuls is starting to come into focus. Norte faces tough competition. Belgium, for instance, has fielded the Dardenne brothers’ neorealist work, Two Day, One Night, while Canada has chosen Xavier Dolan’s Mommy which won a Jury Prize in Cannes this year. Turkey has high hopes for Nuri Bilge Ceylan’s Winter Sleep which beat Mommy in Cannes and took home the coveted Palme d’Or.

After premiering in Cannes last year, Norte enjoyed positive critical reception at nearly all the world’s major festivals including Karlovy Vary, Locarno, Toronto, New York, Busan, Tokyo, Rotterdam and Hong Kong. Whether this is enough to get a slot in the Oscars foreign language film category remains to be seen.

The Philippines has regularly submitted films for Academy Awards consideration since 1953, but none have been nominated.

UNYPAD member awarded international scholarship in Sudan

By Abu Jalil Aiman COTABATO City (September 23, 2014) – Abubakar Salem Abas, Cotabato City Chapter acting Vice President for Internal Affairs of the United Youth for Peace and Development has been awarded scholarship in University of Khartoum, Sudan, Africa. Before his departure, Abas was briefed by some members of the UNYPAD Executive Council with

Probability Theory Problems

Let’s have fun on probability theory, here is my first problem set in the said subject.


  1. It was noted that statisticians who follow the deFinetti school do not accept the Axiom of Countable Additivity, instead adhering to the Axiom of Finite Additivity.
    1. Show that the Axiom of Countable Additivity implies Finite Additivity.
    2. Although, by itself, the Axiom of Finite Additivity does not imply Countable Additivity, suppose we supplement it with the following. Let $A_1supset A_2supsetcdotssupset A_nsupset cdots$ be an infinite sequence of nested sets whose limit is the empty set, which we denote by $A_ndownarrowemptyset$. Consider the following:
      Axiom of Continuity: If $A_ndownarrowemptyset$, then $P(A_n)rightarrow 0$

      Prove that the Axiom of Continuity and the Axiom of Finite Additivity imply Countable Additivity.

  2. Prove each of the following statements. (Assume that any conditioning event has positive probability.)
    1. If $P(B)=1$, then $P(A|B)=P(A)$ for any $A$.
    2. If $Asubset B$, then $P(B|A)=1$ and $P(A|B)=P(A)/P(B)$.
    3. If $A$ and $B$ are mutually exclusive, then begin{equation}nonumber P(A|Acup B) = displaystylefrac{P(A)}{P(A)+P(B)}. end{equation}
    4. $P(Acap Bcap C)=P(A|Bcap C)P(B|C)P(C)$.
  3. Prove that the following functions are cdfs.
    1. $frac{1}{2}+frac{1}{pi}arctan(x), xin (-infty, infty)$
    2. $(1+e^{-x})^{-1},xin (-infty,infty)$
    3. $e^{-e^{-x}}, xin (-infty, infty)$
    4. $1-e^{-x}, xin (0,infty)$
    5. the function defined in (1.5.6), (Check in the reference below.)
  4. A cdf $F_X$ is stochastically greater than a cdf $F_{Y}$ if $F_{X}(t)leq F_{Y}(t)$ for all $t$ and $F_{X}(t) < F_{Y}(t)$ for some $t$. Prove that if $Xsim F_X$ and $Ysim F_Y$, then begin{equation}nonumber P(X>t) geq P(Y>t);text{for every};t end{equation} and begin{equation}nonumber P(X>t)>P(Y>t),;text{for some}; t end{equation} that is, $X$ tends to be bigger than $Y$.
  5. Let $X$ be a continuous random variable with pdf $f(x)$ and cdf $F(x)$. For a fixed number $x_0$, define the function begin{equation}nonumber g(x) = begin{cases} f(x) / [1-F(x_0)]& x geq x_0\ 0 & x < x_0. end{cases} end{equation} Prove that $g(x)$ is a pdf. (Assume that $F(x_0)
  6. For each of the following, determine the value of $c$ that makes $f(x)$ a pdf.
    1. $f(x)=mathrm{c}sin x, 0 < x < pi/2$
    2. $f(x)=mathrm{c}e^{-|x|},-infty < x < infty$


    1. Proof. Let $mathscr{B}$ be a $sigma$-algebra and consider $A_1,A_2,cdotsin mathscr{B}$ are pairwise disjoint, then by countable additivity begin{equation}nonumber Pleft(displaystylebigcup_{i=1}^{infty}A_iright)=displaystylesum_{i=1}^{infty}P(A_i). end{equation} Now, begin{equation} begin{aligned} Pleft(displaystylebigcup_{i=1}^{infty}A_iright)&= Pleft(displaystylebigcup_{i=1}^{n}A_icupdisplaystyle bigcup_{i=n+1}^{infty}A_iright)\ &= Pleft(displaystylebigcup_{i=1}^{n}A_iright)+Pleft(displaystyle bigcup_{i=n+1}^{infty}A_iright),;(text{since};A_i’s;text{are disjoints})\ &=P(A_1)+cdots+P(A_n)+Pleft(displaystyle bigcup_{i=n+1}^{infty}A_iright),\ &quad(text{by finite additivity})\ &=displaystylesum_{i=1}^{n}P(A_i)+Pleft(displaystyle bigcup_{i=n+1}^{infty}A_iright) end{aligned}nonumber end{equation} Notice that for any $n$, we can consider $P(A_i),;i>n$ to be empty. Implying begin{equation}nonumber Pleft(displaystylebigcup_{i=n+1}^{infty}A_iright)=displaystyle sum_{i=n+1}^{infty}P(A_i)=P(emptyset)+P(emptyset)+cdots, end{equation} that is, begin{equation}nonumber begin{aligned} Pleft(displaystylebigcup_{i=1}^{infty}A_iright)&= displaystylesum_{i=1}^{n}P(A_i)+sum_{i=n+1}^{infty}P(A_i)\ &=displaystylesum_{i=1}^{n}P(A_i)+P(emptyset)+P(emptyset)+cdots end{aligned} end{equation} $therefore$ countable additivity implies finite additivity.
    2. From (a), we have shown that countable additivity implies finite additivity, i.e., begin{equation} Pleft(displaystylebigcup_{i=1}^{infty}A_iright)=displaystylesum_{i=1}^{n}P(A_i)+Pleft(displaystyle bigcup_{i=n+1}^{infty}A_iright) nonumber end{equation} If we supplement this with the following condition, that $A_1supset A_2supset A_3supsetcdots$. By Axiom of Continuity, $displaystylelim_{nto infty}A_n=emptyset$, and by Monotone Sequential Continuity, $Pleft(displaystylelim_{ntoinfty}A_nright)= displaystylelim_{ntoinfty}P(A_n)=0$. Now we can write $A_1supset A_2supset A_3supsetcdots$ as begin{equation}nonumber B_k=bigcup_{i=k}^{infty}A_i,;text{such that};B_{k+1}subset B_k, text{implying}; lim_{ktoinfty}B_k=emptyset end{equation} Thus, finite additivity plus axiom of continuity, we have begin{equation}nonumber begin{aligned} Pleft(bigcup_{i=1}^{infty}A_iright)&=lim_{ntoinfty}left( sum_{i=1}^{n}P(A_i)+P(B_{n+1})right)\ &=lim_{ntoinfty}left(sum_{i=1}^{n}P(A_i)right)+lim_{ntoinfty} P(B_{n+1})\ &=sum_{i=1}^{infty}P(A_i)+0,;(text{by axiom of continuity}). end{aligned} end{equation} Implying countable additivity.
    1. Proof. If $P(B)=1$, then $P(S)=P(B)=1$. Because $Asubseteq S$, implies $Asubseteq B$. Thus, $Acap B = A$, and therefore begin{equation}nonumber P(A|B)=displaystylefrac{P(Acap B)}{P(B)}=displaystylefrac{P(A)}{P(B)}=P(A) end{equation} $hspace{12.5cm}blacksquare$
    2. Proof. If $Asubseteq B$ then begin{equation}nonumber P(B|A)=displaystylefrac{P(Acap B)}{P(A)}=displaystylefrac{P(A)}{P(A)}=1 end{equation} and, begin{equation}nonumber P(A|B)=displaystylefrac{P(Acap B)}{P(B)}=displaystylefrac{P(A)}{P(B)} end{equation} $hspace{12.5cm}blacksquare$
    3. Proof. If $A$ and $B$ are mutually exclusive, then begin{equation} nonumber begin{aligned} P(A|Acup B)&=displaystylefrac{P(Acap (Acup B))}{P(Acup B)}\ &=displaystylefrac{P(A)cup [P(Acap B)]}{P(A)+ P(B)}\ &=displaystylefrac{P(A)}{P(A)+ P(B)} end{aligned} end{equation}$hspace{12.5cm}blacksquare$
    4. Proof. Consider, begin{equation}nonumber P(A|Bcap C)=displaystylefrac{P(Acap Bcap C)}{P(Bcap C)} end{equation} Hence, begin{equation}nonumber P(Acap Bcap C) = P(A|Bcap C)P(Bcap C) end{equation} Now $P(Bcap C)=P(B|C)P(C)$, therefore begin{equation}nonumber P(Acap Bcap C) = P(A|Bcap C)P(B|C)P(C) end{equation}$hspace{12.5cm}blacksquare$
  1. $F(x)$ is a cdf if it satisfies the following conditions:
    1. $displaystylelim_{xto-infty}F(x)=0$ and $displaystylelim_{xtoinfty}F(x)=1$
    2. $F(x)$ is nondecreasing.
    3. $F(x)$ is right-continuous.
    1. Proof.
      1. $F(x)=frac{1}{2}+frac{1}{pi}arctan(x), xin (-infty, infty)$

        Above figure was generated by the following $mathrm{LaTeX}$ codes:

        begin{equation}nonumber begin{aligned} displaystylelim_{xto-infty}F(x)&=displaystylelim_{xto-infty} left(frac{1}{2}+frac{1}{pi}arctan(x)right)\ &=frac{1}{2}+frac{1}{pi}displaystylelim_{xto-infty}left(arctan(x)right)\ &=frac{1}{2}+frac{1}{pi} left(frac{-pi}{2}right),;text{since};displaystylelim_{xto-frac{pi}{2}}frac{sin(x)}{cos(x)}=-infty\ &=0\[0.5cm] displaystylelim_{xtoinfty}F(x)&=displaystylelim_{xtoinfty} left(frac{1}{2}+frac{1}{pi}arctan(x)right)\ &=frac{1}{2}+frac{1}{pi}displaystylelim_{xtoinfty}left(arctan(x)right)\ &=frac{1}{2}+frac{1}{pi} left(frac{pi}{2}right),;text{since};displaystylelim_{xtofrac{pi}{2}}frac{sin(x)}{cos(x)}=infty\ &=1 end{aligned} end{equation}

      2. To test if $F(x)$ is nondecreasing, recall in Calculus that, first differentiation of the function tells us if it is decreasing or increasing. In particular, $frac{dF(x)}{dx}>0$ tells us that the function is increasing in a given interval of $x$. Thus, begin{equation} nonumber frac{dF(x)}{dx}=frac{d}{dx}left(frac{1}{2}+frac{1}{pi}arctan(x)right)=frac{1}{pi(1+x^2)} end{equation} Confirm the above differentiation with Python using sympy module.

        Since $x^2$ is always positive for all $x$, thus $frac{dF(x)}{dx}>0$, implying $F(x)$ is increasing.

      3. $F(x)$ is continuous, implies that $F(x)$ is right-continuous.


    2. Proof.
      1. $ F(x)=displaystylefrac{1}{1+e^{-x}}, xin(-infty,infty) $

        begin{equation}nonumber begin{aligned} displaystylelim_{xto-infty}F(x)&=displaystylelim_{xto-infty} left(frac{1}{1+e^{-x}}right)\ &=0\[0.5cm] displaystylelim_{xtoinfty}F(x)&=displaystylelim_{xtoinfty} left(frac{1}{1+e^{-x}}right)\ &=displaystylelim_{xtoinfty} left(frac{1}{1+frac{1}{e^{x}}}right)\ &=1 end{aligned} end{equation} Confirm these in Python,

      2. Using the same method we did in (a), we have begin{equation} nonumber begin{aligned} frac{dF(x)}{dx}&=frac{d}{dx}left(displaystylefrac{1}{1+e^{-x}}right)\ &=frac{e^{-x}}{(1+e^{-x})^2} end{aligned} end{equation} $frac{dF(x)}{dx}=frac{e^{-x}}{(1+e^{-x})^2}>0,;forall;xin(-infty,infty)$. Thus the function is increasing in the interval of $x$.
      3. $F(x)$ is continuous, implies the function is right-continuous.


    3. Proof.
      1. $F(x)=e^{-e^{-x}}, xin (-infty, infty)$

        begin{equation}nonumber begin{aligned} displaystylelim_{xto-infty}F(x)&=displaystylelim_{xto-infty} left(e^{-e^{-x}}right)\ &=displaystylelim_{xto-infty} left(frac{1}{e^{frac{1}{e^{x}}}}right)\ &=0\[0.5cm] displaystylelim_{xtoinfty}F(x)&=displaystylelim_{xtoinfty} left(e^{-e^{-x}}right)\ &=displaystylelim_{xtoinfty} left(frac{1}{e^{frac{1}{e^{x}}}}right)\ &=1 end{aligned} end{equation}

      2. Like what we did above, $frac{dF(x)}{dx}$ is, begin{equation} nonumber frac{dF(x)}{dx}=frac{d}{dx}left(e^{-e^{-x}}right)=e^{-x}e^{-e^{-x}}>0 end{equation} Because $e^{-x}e^{-e^{-x}}>0,;forall; xin(-infty,infty)$. Then we say $F(x)$ is an increasing function in the interval of $x$.
      3. $F(x)$ is continuous, implies that $F(x)$ is right-continuous.


    4. Proof.
      1. $F(x)=1-displaystylefrac{1}{e^{x}}, xin(0,infty)$

        begin{equation}nonumber begin{aligned} displaystylelim_{xto-infty}F(x)&=displaystyle lim_{xto 0}F(x)=1-displaystylelim_{xto 0} left(frac{1}{e^{x}}right) =0\[0.5cm] displaystylelim_{xtoinfty}F(x)&=1- displaystylelim_{xtoinfty} left(frac{1}{e^{x}}right)=1 end{aligned} end{equation}

      2. begin{equation}nonumber frac{dF(x)}{dx}=frac{d}{dx}left(1-frac{1}{e^{x}}right)=0-(-e^{-x})=frac{1}{e^{x}} end{equation} $F(x)$ is an increasing function since $frac{1}{e^{x}}>0,;forall;xin(0,infty)$.
      3. $F(x)$ is right-continuous, since it is continuous.


    5. Proof. The function in Equation (1.5.6) is given by, begin{equation} F_Y(y)=begin{cases} displaystylefrac{1-varepsilon}{1+e^{-y}}&text{if};yvarepsilon>0\ varepsilon+displaystylefrac{1-varepsilon}{1+e^{-y}}&text{if};ygeq 0,;text{for some};varepsilon, 1>varepsilon>0 end{cases}nonumber end{equation}
      1. begin{equation}nonumber begin{aligned} displaystylelim_{yto-infty}F_Y(y)&=displaystylelim_{yto-infty} left(displaystylefrac{1-varepsilon}{1+e^{-y}}right)=displaystylelim_{yto-infty} left(displaystylefrac{1-varepsilon}{1+frac{1}{e^{y}}}right)=0\[0.5cm] displaystylelim_{ytoinfty}F(y)&=displaystylelim_{ytoinfty} left(varepsilon+displaystylefrac{1-varepsilon}{1+e^{-y}}right)=varepsilon + displaystylelim_{ytoinfty} left(displaystylefrac{1-varepsilon}{1+frac{1}{e^{y}}}right)=1 end{aligned} end{equation}
      2. For $y0$ since $00$, implying that the function is increasing.

        For $ygeq 0$, begin{equation} begin{aligned} frac{d}{dy}left(varepsilon+frac{1-varepsilon}{1+e^{-y}}right)&=varepsilon+frac{(1-varepsilon)e^{-y}}{(1+e^{-y})^2} end{aligned}nonumber end{equation} The function is increasing since $varepsilon + frac{(1-varepsilon)e^{-y}}{(1+e^{-y})^2}>0$ for all $ygeq 0$.

      3. Since the function is continuous, then the function is right-continuous.


  2. Proof. We know that, begin{equation}nonumber P(X>t)=1-P(Xleq t)=1-F_X(t) end{equation} and begin{equation}nonumber P(Y>t)=1-P(Yleq t)=1-F_Y(t) end{equation} Hence we have, begin{equation}nonumber begin{aligned} P(X>t)=1-F_X(t);&overset{?}{geq};1-F_Y(t)=P(Y>t)\ end{aligned} end{equation} Since $F_X(t)leq F_Y(t)$, then the difference $1-F_X(t)$ tends to get bigger than $1-F_Y(t)$. Thus for all $t$, $P(X>t)geq P(X>t)$.

    Now if $F_X(t) < F_Y(t)$ for some $t$, then using the same argument above, $P(X>t) > P(X>t)$ for some $t$.

  3. Proof. For a function to be a pdf, it has to satisfy the following:
    1. $g(x)geq 0$ for all $x$; and,
    2. $displaystyleint_{-infty}^{infty}g(x),dx=1$.

    For any arbitrary $x_0$, $F(x_0)

  4. In order for $f(x)$ to be a pdf, it has to integrate to 1.
    1. begin{equation} begin{aligned} int_{-infty}^{infty}f(x)&=int_{0}^{frac{pi}{2}}mathrm{c}sin x=displaystyleleft.-(mathrm{c})cos xdisplaystylerightrvert_{0}^{frac{pi}{2}}\ &=-mathrm{c}left(cosleft(frac{pi}{2}right)-cos(0)right)\ &=-mathrm{c}(0-1)=1mathrm{c} end{aligned}nonumber end{equation} Hence, $mathrm{c}$ is 1. Confirm this with python,

    2. begin{equation} begin{aligned} int_{-infty}^{infty}f(x)&=int_{-infty}^{infty} mathrm{c},e^{-|x|}\ &=mathrm{c}left(int_{-infty}^{0} ,e^{x},dx+int_{0}^{infty} e^{-x},dxright)\ &=mathrm{c}left[(e^{0}-e^{-infty})-(e^{-infty}-e^{0})right]\ &=mathrm{c}(1+1) = 2mathrm{c} end{aligned}nonumber end{equation} Hence, c is $frac{1}{2}$. Confirm this with Python,


  1. Casella, G. and Berger, R.L. (2001). Statistical Inference. Thomson Learning, Inc.

salamindanaw returns

The second edition of SalaMindanaw International Film Festival will take place from November 25 to 29, 2014.The call for submissions is now open for the following sections: Asian full length, Asian shorts, and Mindanao shorts. Narrative, documentary an…

the birth of tradition

This is a behind the scene photo of the first scene to be shot in The Obscured Histories and Silent Longings of Daguluan’s Children in Matanao, Davao del Sur, in 2009.  We shot the film without a script.  We went on location with a big idea, …


is a mantra you learn in film school. It’s the Om Namma Shivaya of the indie movement when a studio to bankroll your film project is hard to come by. When you’re a student filmmaker, the next best thing to a Harvey Weinstein is Mom and Dad. You suddenly become extra sweet to them. You become subservient to their every command. You wash the dishes, scrub the bathroom tiles, or take out the garbage. The maid you explain has too many errands on her hands. She can use some help around the house. Besides you have free time. Retreating to your room you suppress the urge to squeal like a pig. You maintain your calm. You assure yourself that patience is a virtue. Later you can throw tantrums at your production team and make an awful excuse for this behavior by saying, “I am an artist. I am entitled to my fits.”
After school your parents expect you to PA for, or if you’re truly lucky and talented, write or direct the next blockbuster starring Daniel and Kathryn. But you haughtily dismiss this idea saying that it’s beneath you. Dad shakes his head and regrets supporting your choice to go to film school. Deep in his heart, he still wants you to have a change of heart and become a bureaucrat or a lawyer.
Finally when you have an idea for your debut film –a ten-hour movie composed of a dozen shots featuring a cart, a horse, a transvestite impersonating the Virgen de la Guadalupe, and an awfully bored peasant reflecting on a koan— your father reluctantly gives you money to produce it. It’s the rave in European art houses you explain. You appeal to his guilt.
“Dad, I would only get half of your money. Don’t worry. I already wrote a film proposal to (insert name of a funding organization).”
As soon as Dad leaves the room, you revert to the saccharine personality of an eight-year-old kid and ask Mom if she can cook for ten members of the production, three meals a day for two weeks. Before she can even protest you shower her profusely with gratitude, sealing your contrived act with a kiss on her forehead.   
Then you talk to other potential producers. Dilemma. Should you wear the signature indie-filmmaker look – denim jeans, ukay-ukay shirt, and Chucks? This way you will look sincere. Or show off in the latest Italian designer shirt and jeans? It can give a semblance of financial stability. The downside is you cannot beg because you will only look like a poseur.
When the potential producer starts quoting the Far Eastern Economic Review in an attempt to turn you down without hurting your feelings, you proceed to Plan B.
You post messages on various art and film e-groups.
In need of actors and camera!
Minimal fee. Independent production.
The truth is you don’t have a budget for camera rental. You only intend to borrow. You can repay this favor by being the PA of the camera’s owner when he makes his film.

At a small café, you nurse a cup of Americano during a nightlong chat with critic friends, the same people you expect to canonize you as the messiah of Philippine cinema when your rough cut premieres in a festival barely three months away. You profess gratitude to advances in digital technology because it’s now possible for you to shoot on HD which is so much cheaper than celluloid. The audience won’t know the difference. Otherwise, your entire production budget will be spent on 35mm’s and film processing.
You attend film festivals and forums. You chat with everybody whom you think can help you realize your dream. You pimp yourself and your project. Then a bald and obese producer who claims to have worked in Hollywood takes interest in your project. He invites you to discuss the project at length in his hotel room. You say yes only to find out later, twenty floors above ground, overlooking the Manila skyline on an early Thursday evening that the producer only wants to bed you. Suddenly you remember that Visconti’s Death in Venice is showing in ten minutes at the mall. You make lame excuses and dart outside the room.
You walk hurriedly to the mall. You become desperate. Your opus may never be produced. Then you hear Darth Vader’s theme in your head. You realize that the dark side is not such a bad idea. You follow Anakin’s steps.
You pause on EDSA and Vargas, suck in the thick polluted air of the metropolis, and regret leaving the producer in his hotel room. You want to go back to his room. You can engage him in a tequila-drinking match – one contest which you always win. Once he is down you can run with the camera and boom sitting nicely on the table. You assure yourself it’s achievable. After all, your film teacher did the same thing to produce his first opus.

(I wrote this blog entry many years ago. Since I’ve been conducting film workshops recently and will be teaching in another one this coming October, I think it’s something worth reading for my workshop students so that they can reflect on whether or not they really want to become filmmakers.)


Four days ago, I was riding a tricycle with my friend Ryan in Zamboanga. He asked me how did I develop my literary writing when all these years I’ve worked as a technical writer or journalist. The former is fluid and personal, while the latter has to a…

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PHOTOS: It’s Showtime host Jhong Hilario thrills a fan in Bukidnon

The post PHOTOS: It’s Showtime host Jhong Hilario thrills a fan in Bukidnon appeared first on Mindanaoan | @mindanaoan.

“It’s Showtime” hurado Jhong Hilario was one of the celebrities who visited Bukidnon for the 2014 Kaamulan Festival, the yearly gathering of the Province of Bukidnon’s 7 hill tribes. Jhong Hilario joined Kamikazee for the SMART sponsored night held last August 31, 2014 at the Capitol Grounds, Malaybalay City. Thank …

The post PHOTOS: It’s Showtime host Jhong Hilario thrills a fan in Bukidnon appeared first on Mindanaoan | @mindanaoan.

PHOTOS: Kapuso celebrities during Kaamulan Festival 2014 in Bukidnon

The post PHOTOS: Kapuso celebrities during Kaamulan Festival 2014 in Bukidnon appeared first on Mindanaoan | @mindanaoan.

If ABS CBN brought in Kathryn Bernardo and Daniel Padilla for Kaamulan Festival 2014 (Bukidnon’s yearly gathering of 7 hill tribes), GMA 7 brought in 4 of their celebrities. During the Kapuso Night held last August 30, 2014 at the Capitol Grounds in Malaybalay City, Yasmien Kurdi, JC Tiuseco, Enzo …

The post PHOTOS: Kapuso celebrities during Kaamulan Festival 2014 in Bukidnon appeared first on Mindanaoan | @mindanaoan.

The ‘Group Picture’

by Carolyn Villamarzo-Juanday   I was sorting out my old school files when I came across this group picture taken pretty long-years ago. When I looked at the faces of those in the photo, I can’t help but reminisce the days when I was still part of this big family. I miss the moment working […]

Highly Recommended Degree Program on Islamic Studies Worldwide

For those who would like to study a degree program on Islamic Studies, here are my highly recommended specialized/higher learning institutions where ijtihad (independent thinking) and academic freedom are commonly practiced:

1. University of Chica…

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CORDAID partner NGOs undergo training on Basic Photography and Success Story Writing

Mindanao Convergence of NGOs for Social Protection (MinConSP) conducted a training- workshop on photography and success story writing on August 18-20 at Eco Resort, Lake Sebu South Cotabato. Twenty two representatives from the Convergence network took part in the said workshop with Bobby Timonera and Marcos Mordeno of Mindanews as trainers-resource persons. The workshop was […]

KFI trains new ANAK KAWAGIB members on advocacy and lobbying skills

Seeing the need to develop a new generation of young Bangsamoro advocates and activists for children’s rights, KFI conducted a training on advocacy and lobbying for 25 new members of ANAK KAWAGIB at Paglat National High School last August 21, 2014. The main aim of the said training was to enhance the capacities of the […]

Lebesgue Measure and Outer Measure Problems

More proving, still on Real Analysis. This is my solution and if you find any errors, do let me know.


Lebesgue Measure: Let $mu$ be set function defined for all set in $sigma$-algebra $mathscr{F}$ with values in $[0,infty]$. Assume $mu$ is countably additive over countable disjoint collections of sets in $mathscr{F}$.

  1. Prove that if $A$ and $B$ are two sets in $mathscr{F}$, with $Asubseteq B$, then $mu(A)leq mu(B)$. This property is called monotonicity.
  2. Prove that if there is a set $A$ in the collection $mathscr{F}$ for which $mu(A)<infty$, then $mu(emptyset)=0$.
  3. Let ${E_{k}}_{k=1}^{infty}$ be a countable collection of sets in $mathscr{F}$. Prove that $muleft(displaystylebigcup_{k=1}^{infty}E_{k}right)leq displaystylesum_{k=1}^{infty}mu(E_k)$

Lebesgue Outer Measure:

  1. By using property of outer measure, prove that the interval $[0,1]$ is not countable.
  2. Let $A$ be the set of irrational numbers in the interval $[0,1]$. Prove that $mu^{*}(A)=1$.
  3. Let $B$ be the set of rational numbers in the interval $[0,1]$, and let ${I_k}_{k=1}^{n}$ be finite collection of open intervals that covers $B$. Prove that $displaystylesum_{k=1}^{n}mu^{*}(I_k)geq 1$.
  4. Prove that if $mu^{*}(A)=0$, then $mu^{*}(Acup B)=mu^{*}(B).$


  1. Proof. If $Asubseteq B$, then $B= Acup (Bcap A^c)Rightarrow B= Acup (Bbackslash A)$. Thus, begin{equation}nonumber begin{aligned} mu(B)&= mu(Acup (Bbackslash A))\ &= mu(A)+mu(Bbackslash A)\ &(text{since};mu;text{is countably additive on disjoint sets}) end{aligned} end{equation} We can see that $mu(B)geq mu(A)$ since $mu(Bbackslash A) > 0$. $hspace{13.5cm}blacksquare$
  2. Proof. For any set $A$ in $mathscr{F}$ such that $mu(A)
  3. Proof. We define a sequence ${A_n}_{n=1}^{infty}subseteqmathscr{F}$, such that $A_1=E_1$ and begin{equation}nonumber A_n = E_n backslash bigcup_{k=1}^{n-1}E_k,;text{for};n>1 end{equation} It is easy to see that $A_n$ is pairwise disjoint, and $bigcup_{n=1}^{infty}A_n=bigcup_{k=1}^{infty}E_k$, also ${A_n}subseteq {E_k}$. Thus by countably additive and monotonicity property of $mu$, we have begin{equation}nonumber begin{aligned} muleft(bigcup_{k=1}^{infty}E_kright)&=muleft(bigcup_{n=1}^{infty}A_nright)\ &=sum_{n=1}^{infty}mu(A_n)\ &leq sum_{k=1}^{infty}mu(E_k);(text{by monotonicty}). end{aligned} end{equation} $hspace{13.5cm}blacksquare$
  4. Proof. Let’s prove this by contradiction, assume the interval $[0,1]$ is countable. Then we need to show that $mu^{*}([0,1])=0$ for it to be countable. Now consider $varepsilon >0$, such that $I = {[varepsilon – 0, 1 + varepsilon]}$ covers $[0,1]$. Then by property of outer measure that says, $mu^{*}([a,b])$ is the length of $[a,b]$, we have begin{equation}nonumber mu^{*}([0,1]) = inf,{ell (I)} = (1+varepsilon) – (0-varepsilon) = 1+2varepsilon end{equation} This holds for each $varepsilon >0$, thus $mu^{*}([0,1])=1$ which is a contradiction.$hspace{2.13cm}blacksquare$
  5. Proof. If $A={mathbb{Q}^ccap [0,1]}$ is the set of irrational numbers in the interval $[0,1]$, then $A^c={mathbb{Q}cap [0,1]}$ is the set of rational numbers in the interval $[0,1]$. Now consider the following, begin{equation}nonumber begin{aligned} mu^{*}([0,1])&=mu^{*}(A)+mu^{*}(A^{c})\ mu^{*}(A)&=mu^{*}([0,1]) – mu^{*}(A^{c})\ &=1 -mu^{*}(A^{c}) end{aligned} end{equation} We need to show that $mu^{*}(A^{c})$ has outer measure zero. To do that, let $a_1,a_2,cdotsin A^{c}$. And for $varepsilon > 0$, $exists$ ${I_n}$ such that ${I_n} = {(a_n – frac{varepsilon}{2^{k+1}}, a_n + frac{varepsilon}{2^{k+1}})}$. Thus, $bigcup_{n=1}^{infty}I_n$ covers $A^{c}$, and by outer measure we have, begin{equation}nonumber begin{aligned} mu^{*}(A^c)& leq infleft{sum_{n=1}^{infty}ell(I_n)right}\ &leq infleft{sum_{n=1}^{infty}left(a_n + frac{varepsilon}{2^{k+1}} – a_n + frac{varepsilon}{2^{k+1}}right) right}\ &leq infleft{sum_{n=1}^{infty}left(frac{varepsilon}{2^{k}}right) right}\ &leq varepsilon end{aligned} end{equation} Since this hold for each $varepsilon$ then $mu^{*}(A^c)=0$. Thus, $mu^{*}(A)=1-0=1$.
  6. Proof. The rational numbers are dense in $mathbb{R}$. Thus, any point in the interval $[0,1]$ may it be irrational numbers will always be a point of closure of $B$, that is $bar{B}=[0,1]$. Since $Bsubseteq bigcup_{k=1}^{infty}I_k$, then by closure property, $bar{B}subseteq overline{bigcup_{k=1}^{infty}I_k}$, which is $bar{B}subseteq bigcup_{k=1}^{infty}bar{I}_k$. Thus by definition of outer measure we have, begin{equation}nonumber begin{aligned} 1=mu^{*}([0,1])&=mu^{*}(bar{B})leq mu^{*}left(bigcup_{k=1}^{infty}bar{I}_kright)\ &leq sum_{k=1}^{infty}mu^{*}(bar{I}_k)=sum_{k=1}^{infty}mu^{*}(I_k). end{aligned} end{equation} Thus, begin{equation}nonumbersum_{k=1}^{infty}mu^{*}(I_k)geq 1 end{equation} $hspace{13.5cm}blacksquare$
  7. Proof. We need to show that, begin{equation}nonumber begin{aligned} &mu^{*}(Acup B)leq mu^{*}(B)\ &mu^{*}(B)leq mu^{*}(Acup B) end{aligned} end{equation}
    1. From the definition of outer measure, begin{equation} nonumber begin{aligned} mu^{*}(Acup B)&leq mu^{*}(A)+mu^{*}(B)\ &leq mu^{*}(B). end{aligned} end{equation}
    2. Since $Bsubseteq Acup B$, then from property of outer measure that if $Asubseteq B$, then $mu^{*}(A)leq mu^{*}(B)$. Hence, begin{equation}nonumber mu^{*}(B)leq mu^{*}(Acup B) end{equation}



  1. Royden, H.L. and Fitzpatrick, P.M. (2010). Real Analysis. Pearson Education, Inc.

PHOTOS: Daniel Padilla and Kathryn Bernardo in Bukidnon #KathNiel

The post PHOTOS: Daniel Padilla and Kathryn Bernardo in Bukidnon #KathNiel appeared first on Mindanaoan | @mindanaoan.

The Province of Bukidnon just finished celebrating its 100th founding anniversary last September 1, 2014. The province capped the celebration with a pyromusical display and a special concert featuring popular comedian Vice Ganda and top loveteam KathNiel – Kathryn Bernardo and Daniel Padilla. An estimated 300,000 strong crowd gathered at …

The post PHOTOS: Daniel Padilla and Kathryn Bernardo in Bukidnon #KathNiel appeared first on Mindanaoan | @mindanaoan.