Monthly Archives: May 2015

Definitely Dahican.

When you’re planning to go on a trip, make sure to indulge yourself and live to that moment. As what others would say “YOLO”, make the most out of it. Experience is one of the best things especially when you see new things. The common reasons why people go to the beach is of course […]

Make Your Own Havaianas 2015 CDO #MYOHCDO2015

The Make Your Own Havaianas 2015 CDO leg was one of the fashion events I looked forward to attending this year. I just love attending MYOH events since I like customizing my very own Havaianas by choosing my preferred straps, soles and special pins. He…

Surviving Crazy, Beautiful Boracay

When I was asked if I wanted to experience Labor Day 2015 in Boracay Island (no less), of course my answer was a resounding YES! Although it wasn’t my first time in an island recognized as one of the world’s finest, it was my first LaBoracay trip and the mere idea of partying with a sea of people piqued my curiosity. And so, armed with my travel essentials, I took two plane rides, a bus ride and a boat outrigger ride to reach the famed Boracay. The agenda was very simple: take over summer! Enjoy the island and what it

bruce

This is Bruce, one of the eight cats that cohabit a two-room apartment with me and and my partner Mogley. One month ago, Bruce underwent a surgery to remove stones from his bladder. He’s doing fine thanks to prayers from friends and family and a lot of…

US Embassy Manila donates Gibbons guitar to Liceo de Cagayan

A school in Cagayan de Oro, Mindanao recently became the recipient of a very special, limited edition Billy Gibbons guitar. The U.S. Embassy Manila donated a Billy Gibbons electric guitar to Liceo de Cagayan University College of Music. This CDO university was chosen to receive the guitar because of the University’s commitment to serve their music students through scholarship programs. Representing the U.S. Embassy was Cultural Affairs Officer Kristin Kneedler who officially donated the guitar. William Frederick “Billy” Gibbons is an American musician. Billy Gibbons is best known as the guitarist of the rock band ZZ Top. Gibbons was ranked

The Happy List #5

Photo from http://www.sketchesofmind.com/
For the past two months my family and I have been through a lot and it seems that it drained almost all the positivity in me and that sadness have succumbed me. And I think it’s about time that i’ll do my happy list because listing down the simplest joys – I believe can bring back some positivity in my life. 

“But I believe good things happen everyday. 
I believe good things happen even when bad things happen.
And I believe on a happy day like today, we can still feel a little sad.
And that’s life, isn’t it?”
-Elsewhere by Gabrielle Zevin

    • My husband Ali finally home after 8 months of staying in Kuwait.
    • Quality time with the husband.
    • Moving my blog to its new home.
    • Knowing that at the end of the day blood is still thicker than water.
    • Cheap scarves! 
    • Lazy days at home.
    • Authentic kebab that tastes heavenly.
    • Cooking sessions with the husband.
    • Colleen Hoover books!
    • Seeing my succulents and cacti bloom. Finally!
    • Food dates with Mariel.
    • Spending an afternoon with my bestie – Jessa.
    • Landing a job that i’m passionate about. My first real job.
    • Being thankful every morning that God added another day into my life.

    Join the Happy List link-up by Camie Juan now! 🙂

    UNYPAD Conducts Peace Forum

    By: Jordan Talusob The United Youth for Peace and Development, Inc. (UNYPAD) conducted a Peace Forum  with theme, “From Boarders to Bridges: Efforts for Inclusive Peace” on May 22, 2015 at the Bangsamoro Leadership and Management Institute (BLMI) Training Center, Simuay, Sultan Kudarat, Maguindanao. The Kroc Institute for International Peace Studies students led by Christine

    the maguindanaon realm, a context

    Excerpt from the manuscript of Notes of Bantugan’s Last Journey which will be out soon.
    The Maguindanaon people—the largest ethnolinguistic group in Central Mindanao—occupy the basin of the Pulangi River (Rio Grande de Mindanao in Spanish accounts), which spills down the southern slopes of the Bukidnon massif in north-central Mindanao, snaking south and west across a low-lying marshy plain to Illana Bay[i]. About twenty miles before reaching the sea the Pulangi splits into two branches. The narrower southern fork is known as the Tamontaka River. Near the mouth of the river, stands Timako Hill, and offshore, the dark crescent of Bongo Island. The wider north fork of the Pulangi flows past Cotabato City, which is located on its south bank about four miles above the river mouth[ii].
    In the past, the Maguindanaon settled along riverbanks and in the valley regions of the Pulangi River where periodic flooding was experienced. It is due to this inundation that the people occupying the area came to be called maguindanaon –“people of the flood plains.” Today, they are found in several provinces. Maguindanao province accounts for 76 percent of the total Maguindanaon population. In Cotabato province, they are concentrated mainly in Pikit, Kabacan, and the interior villages of Midsayap. In Sultan Kudarat province, they live in Lutayan, as well as the coastal towns of Lebak, Kalamansig, and Palembang. They are also found in Malapatan, in Sarangani province, and Dinas and Labangan in Zamboanga Sibugay.
    The Maguindanaon people descended from the waves of Proto- and Deutro Malays migrants from mainland Asia in 3,000 B.C. Exhibiting a higher stage of social development, they formed settlements or communities with political organizations along family or blood lines. Like most descendants of Proto- and Deutro Malays in Southeast Asia, they were animists believing that the rivers, trees, jungles, and mountains were inhabited by malevolent spirits which motivated them to develop elaborate rituals to placate these spirits[iii]. One can surmise that there already existed various forms of ritualistic practices by the time they came in contact with the rest of the region –starting with Indianized state of Funan in 1 A.D., to the contact with Islamic traders and sufis in the late 1500s.
     In the early sixteenth century, Shariff Muhammad Kabungsuan of Johore arrived in Mindanao, landing in what is now Malabang, in Lanao del Sur, and introduced Islam to the native population. He founded the Maguindanao sultanate through his marriage to the daughters of local chieftains. One of his daughters, Mamur, was married to the Buayan chief Pulwa. Thus, the Maguindanao and Buayan sultanate claim descent from Kabungsuan. Later on, succeeding sultanates trace their ancestry from him as well.
    During the course of history, the Buayan rulers controlled the datus and territories in the upper valley of the Pulangi –tau sa laya—while the Maguindanao sultans came to control the lower valley –tau sa ilud. Though interdependent, and related through marital and blood ties, they competed against each other for supremacy for most of their histories.
    The distinction into tau sa laya and tau sa ilud as inland and coastal people has its dangers[iv], but the contrasts between the two is important in this study. Geographic factors and dialect variations also mean differences in ritual practices. Still both have a lot in common. The traditional Maguindanaon were horticulturists, growing either rice in upland fields or wet rice in lowland paddies. In the modern period, they have shifted to plow and harrow method of wet rice cultivation[v]. The Maguindanaon are excellent fishermen in both the riverine areas and coastlines. They possess a strong weaving, metal craft, carving, and musical tradition.
    Although one of the thirteen Muslim ethnolinguistic groups in Mindanao, the Maguindanaon people retained indigenous belief systems in their religious culture. It is a reflection of how Islam is practiced by the vast majority of Muslims in Southeast Asia –moderate, tolerant and syncretic.
    [i] Ileto, Reynaldo, Magindanao 1860-1888 The Career of Datu Utto of Buayan, anvil Publishing, 2007.
    [ii] McKenna, Thomas, Muslim Rulers and Rebels, Anvil Publishing, 1998.
    [iii] Nasuruddin, Mohammad Ghouse, The Malay Traditional Music, Dewan Bahasa dan Pustaka, 1992.
    [iv] Ibid, 1.
    [v] Stewart, James C., People of the Flood Plain, Ph.D. dissertation, Department of Anthropology, University of Hawaii, 1977.

    How I Use Globe GCash MasterCard As Easy As 1-2-3

    If, like me, you’re the type of person who likes the convenience of shopping or paying bills (online or offline) using a debit or credit card BUT requires an easy-to-understand way to keep track of all the spending OR if you need a credit card sans the hassle, then you will love the new Globe GCash MasterCard. I learned the beauty of this baby just a few weeks ago during a trip to the gorgeous, world-famous Boracay Island, where I experienced the whole LaBoracay 2015 madness. I was in Boracay not only to bask in the sun or for the

    New place for prayer and reflection in Malaybalay City Bukidnon

    There’s a new place ideal for prayer and reflection in Bukidnon, Mindanao, Philippines. Q Park II, located at Barangay Kalasungay, Malaybalay City, boasts of a lovely manicured lawn, a gorgeous view of the Bukidnon mountain ranges and Malaybalay’s signature cool breeze. It’s a nice place to commune with the Lord. An alternative to the Monastery of Transfiguration, which is also located in Malaybalay. Admission to Q Park II is FREE and this park is open from 9 AM to 5 PM, Tuesday to Sunday. Closed on Mondays. There are several huts you can use. You can’t miss Q Park II.

    Parametric Inference: Likelihood Ratio Test Problem 2

    More on Likelihood Ratio Test, the following problem is originally from Casella and Berger (2001), exercise 8.12.

    Problem

    For samples of size $n=1,4,16,64,100$ from a normal population with mean $mu$ and known variance $sigma^2$, plot the power function of the following LRTs (Likelihood Ratio Tests). Take $alpha = .05$.

    1. $H_0:muleq 0$ versus $H_1:mu>0$
    2. $H_0:mu=0$ versus $H_1:muneq 0$

    Solution

    1. The LRT statistic is given by $$ lambda(mathbf{x})=frac{displaystylesup_{muleq 0}mathcal{L}(mu|mathbf{x})}{displaystylesup_{-infty<mu<infty}mathcal{L}(mu|mathbf{x})}, ;text{since }sigma^2text{ is known}. $$ The denominator can be expanded as follows: $$ begin{aligned} sup_{-infty<mu<infty}mathcal{L}(mu|mathbf{x})&=sup_{-infty<mu<infty}prod_{i=1}^{n}frac{1}{sqrt{2pi}sigma}expleft[-frac{(x_i-mu)^2}{2sigma^2}right]\ &=sup_{-infty<mu<infty}frac{1}{(2pisigma^2)^{1/n}}expleft[-displaystylesum_{i=1}^{n}frac{(x_i-mu)^2}{2sigma^2}right]\ &=frac{1}{(2pisigma^2)^{1/n}}expleft[-displaystylesum_{i=1}^{n}frac{(x_i-bar{x})^2}{2sigma^2}right],\ &quadtext{since }bar{x}text{ is the MLE of }mu.\ &=frac{1}{(2pisigma^2)^{1/n}}expleft[-frac{n-1}{n-1}displaystylesum_{i=1}^{n}frac{(x_i-bar{x})^2}{2sigma^2}right]\ &=frac{1}{(2pisigma^2)^{1/n}}expleft[-frac{(n-1)s^2}{2sigma^2}right],\ end{aligned} $$ while the numerator is evaluated as follows: $$ begin{aligned} sup_{muleq 0}mathcal{L}(mu|mathbf{x})&=sup_{muleq 0}prod_{i=1}^{n}frac{1}{sqrt{2pi}sigma}expleft[-frac{(x_i-mu)^2}{2sigma^2}right]\ &=sup_{muleq 0}frac{1}{(2pisigma^2)^{1/n}}expleft[-displaystylesum_{i=1}^{n}frac{(x_i-mu)^2}{2sigma^2}right]. end{aligned} $$ Above expression will attain its maximum if the value inside the exponential function is small. And for negative values of $muin(-infty,0)$ the quantity $(x_i-mu)^2$ would be large, implies that the exponential term would become small. Therefore, the only value that will give us the supremum likelihood is $mu=mu_0=0$. Hence, $$ begin{aligned} sup_{muleq 0}mathcal{L}(mu|mathbf{x})&=frac{1}{(2pisigma^2)^{1/n}}expleft[-displaystylesum_{i=1}^{n}frac{(x_i-mu_0)^2}{2sigma^2}right]\ =frac{1}{(2pisigma^2)^{1/n}}&expleft[-displaystylesum_{i=1}^{n}frac{(x_i-bar{x}+bar{x}-mu_0)^2}{2sigma^2}right]\ =frac{1}{(2pisigma^2)^{1/n}}&expleft{-displaystylesum_{i=1}^{n}left[frac{(x_i-bar{x})^2+2(x_i-bar{x})(bar{x}-mu_0)+(bar{x}-mu_0)^2}{2sigma^2}right]right}\ =frac{1}{(2pisigma^2)^{1/n}}&expleft[-frac{(n-1)s^2+n(bar{x}-mu_0)^2}{2sigma^2}right], \ &text{since the middle term is 0.}\ =frac{1}{(2pisigma^2)^{1/n}}&expleft[-frac{(n-1)s^2+nbar{x}^2}{2sigma^2}right], text{since }mu_0=0.\ end{aligned} $$ So that $$ begin{equation} label{eq:lrtre} begin{aligned} lambda(mathbf{x})&=frac{frac{1}{(2pisigma^2)^{1/n}}expleft[-frac{(n-1)s^2+nbar{x}^2}{2sigma^2}right]}{frac{1}{(2pisigma^2)^{1/n}}expleft[-frac{(n-1)s^2}{2sigma^2}right]}\ &=expleft[-frac{nbar{x}^2}{2sigma^2}right].\ end{aligned} end{equation} $$ And we reject the null hypothesis if $lambda(mathbf{x})leq c$, that is $$ begin{aligned} expleft[-frac{nbar{x}^2}{2sigma^2}right]&leq c\ -frac{nbar{x}^2}{2sigma^2}&leq log c\ frac{lvertbar{x}rvert}{sigma/sqrt{n}}&geqsqrt{-2log c}=c’. end{aligned} $$
      Figure 1: Plot of Likelihood Ratio Test Statistic for $n = 4,sigma = 1$.

      Hence, rejecting the null hypothesis if $lambda(mathbf{x})leq c$, is equivalent to rejecting $H_0$ if $frac{bar{x}}{sigma/sqrt{n}}geq c’in[0,infty)$. Figure 1 depicts the plot of the LRT, the shaded region is on the positive side because that’s where the alternative region is, $H_1:mu>0$, in a sense that if the LRT is small enough to reject $H_0$, then it simply tells us that the plausibility of the parameter in the alternative in explaining the sample is higher compared to the null hypothesis. And if that’s the case, we expect the sample to come from the model proposed by $H_1$, so that the sample mean $bar{x}$, being an unbiased estimator of the population mean $mu$, a function of the LRT statistic, should fall on the side (shaded region) of the alternative.

      So that the power function, that is the probability of rejecting the null hypothesis given that it is true (the probability of Type I error) is, $$ begin{aligned} beta(mu)&=mathrm{P}left[frac{bar{x}-mu_0}{sigma/sqrt{n}}geq c’right],quadmu_0=0\ &=1-mathrm{P}left[frac{bar{x}+mu-mu-mu_0}{sigma/sqrt{n}}< c’right]\ &=1-mathrm{P}left[frac{bar{x}-mu}{sigma/sqrt{n}} + frac{mu-mu_0}{sigma/sqrt{n}}< c’right]\ &=1-mathrm{P}left[frac{bar{x}-mu}{sigma/sqrt{n}}< c’+ frac{mu_0-mu}{sigma/sqrt{n}}right]\ &=1-Phileft[c’+ frac{mu_0-mu}{sigma/sqrt{n}}right]. end{aligned} $$ Values taken by $Phi$ are negative and so it decreases, but since we subtracted it to 1, then $beta(mu)$ is an increasing function. So that for $alpha=.05$, $$ begin{aligned} alpha&=sup_{muleq mu_0}beta(mu)\ .05&=beta(mu_0)Rightarrowbeta(mu_0)=1-Phi(c’)\ .95&=Phi(c’)Rightarrow c’=1.645. end{aligned} $$ Since, $$ begin{aligned} Phi(1.645)=int_{-infty}^{1.645}frac{1}{sqrt{2pi}}expleft[-frac{x^2}{2}right]operatorname{d}x=.9500151. end{aligned} $$ Therefore for $c’=1.645,mu_0=0,sigma=1$, the plot of the power function as a function of $mu$ for different sample size, $n$, is shown in Figure 2. For example, for $n=1$ we compute for the function begin{equation} label{eq:powcomp} begin{aligned} beta(mu)&=1-Phileft[c’+ frac{mu_0-mu}{sigma/sqrt{n}}right]\ &=1-Phileft[1.645+ frac{0-mu}{1/sqrt{1}}right]\ &=1-int_{-infty}^{left(1.645+ frac{0-mu}{1/sqrt{1}}right)}frac{1}{sqrt{2pi}}expleft[-frac{x^2}{2}right]operatorname{d}x. end{aligned} end{equation} The obtained values would be the $y$. For $n = 64$ $$ begin{aligned} beta(mu)&=1-Phileft[c’+ frac{mu_0-mu}{sigma/sqrt{n}}right]\ &=1-Phileft[1.645+ frac{0-mu}{1/sqrt{64}}right]\ &=1-int_{-infty}^{left(1.645+ frac{0-mu}{1/sqrt{64}}right)}frac{1}{sqrt{2pi}}expleft[-frac{x^2}{2}right]operatorname{d}x, end{aligned} $$ and so on.

      Figure 2: Power Function for Different Values of $n$.

    2. The LRT statistic is given by $$ lambda(mathbf{x})=frac{displaystylesup_{mu= 0}mathcal{L}(mu|mathbf{x})}{displaystylesup_{-infty
      Figure 3: Plot of Likelihood Ratio Test Statistic for $n = 4,sigma = 1$.

      So that the power function is, $$ begin{aligned} beta(mu)&=mathrm{P}left[frac{lvertbar{x}rvert}{sigma/sqrt{n}}geq c’right]\ &=1 – mathrm{P}left[frac{lvertbar{x}rvert}{sigma/sqrt{n}}
      Figure 4: Two-Sided Power Function for Different $n$.

      The points in the plot are computed by substituting values of $mu=0,sigma=1$ and $n$ to the power function just like we did in Equation (2).

    Reference

    1. Casella, G. and Berger, R.L. (2001). Statistical Inference. Thomson Learning, Inc.
    2. Felix Schönbrodt. Shading regions of the normal: The Stanine scale. Retrieved May 2015.

    the reluctant fellow

    After stalling for the last four years, and with the goading of three senior writers (Vim Nadera, Eugene Evasco and Ricky de Ungria who tried to convince me from August 2012 to February this year), I finally gave in to the urge and submitted an applica…

    Parametric Inference: Likelihood Ratio Test Problem 1

    Another post for mathematical statistics, the problem below is originally from Casella and Berger (2001) (see Reference 1), exercise 8.6.

    Problem

    1. Suppose that we have two independent random samples $X_1,cdots, X_n$ are exponential($theta$), and $Y_1,cdots, Y_m$ are exponential($mu$).
      1. Find the LRT (Likelihood Ratio Test) of $H_0:theta=mu$ versus $H_1:thetaneqmu$.
      2. Show that the test in part (a) can be based on the statistic
      3. $$ T=frac{sum X_i}{sum X_i+sum Y_i}. $$

      4. Find the distribution of $T$ when $H_0$ is true.

    Solution

      1. The Likelihood Ratio Test is given by $$ lambda(mathbf{x},mathbf{y}) = frac{displaystylesup_{theta = mu,mu>0}mathrm{P}(mathbf{x},mathbf{y}|theta,mu)}{displaystylesup_{theta > 0,mu>0}mathrm{P}(mathbf{x}, mathbf{y}|theta,mu)}, $$ where the denominator is evaluated as follows: $$ sup_{theta > 0,mu>0}mathrm{P}(mathbf{x}, mathbf{y}|theta,mu)= sup_{theta > 0}mathrm{P}(mathbf{x}|theta)sup_{mu > 0}mathrm{P}(mathbf{y}|mu),quadtext{by independence.} $$ So that, $$ begin{aligned} sup_{theta > 0}mathrm{P}(mathbf{x}|theta)&=sup_{theta>0}prod_{i=1}^{n}frac{1}{theta}expleft[-frac{x_i}{theta}right]=sup_{theta>0}frac{1}{theta^n}expleft[-frac{sum_{i=1}^{n}x_i}{theta}right]\ &=frac{1}{bar{x}^n}expleft[-frac{sum_{i=1}^{n}x_i}{bar{x}}right]=frac{1}{bar{x}^n}exp[-n], end{aligned} $$ since $bar{x}$, or the sample mean is the MLE of $theta$. Also, $$ begin{aligned} sup_{mu > 0}mathrm{P}(mathbf{y}|mu)&=sup_{mu>0}prod_{j=1}^{m}frac{1}{mu}expleft[-frac{y_j}{mu}right]=sup_{mu>0}frac{1}{mu^m}expleft[-frac{sum_{j=1}^{m}y_j}{mu}right]\ &=frac{1}{bar{y}^m}expleft[-frac{sum_{j=1}^{m}y_j}{bar{y}}right]=frac{1}{bar{y}^m}exp[-m]. end{aligned} $$ Now the numerator is evaluated as follows, $$ begin{aligned} sup_{theta = mu,mu>0}mathrm{P}(mathbf{x},mathbf{y}|theta,mu)&=sup_{theta=mu,mu>0}mathrm{P}(mathbf{x}|theta)mathrm{P}(mathbf{y}|mu),quadtext{by independence.}\ &=sup_{theta=mu,mu>0}prod_{i=1}^{n}frac{1}{theta}expleft[-frac{x_i}{theta}right]prod_{j=1}^{m}frac{1}{mu}expleft[-frac{y_j}{mu}right]\ &=sup_{theta=mu,mu>0}frac{1}{theta^n}expleft[-frac{sum_{i=1}^nx_i}{theta}right]frac{1}{mu^m}expleft[-frac{sum_{j=1}^m y_j}{mu}right]\ &=sup_{mu>0}frac{1}{mu^n}expleft[-frac{sum_{i=1}^nx_i}{mu}right]frac{1}{mu^m}expleft[-frac{sum_{j=1}^m y_j}{mu}right]\ &=sup_{mu>0}frac{1}{mu^{n+m}}expleft{-frac{1}{mu}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]right} end{aligned} $$ Note that $mu$ is a nuisance parameter, and so we will also maximize this over its domain. And to do that we take the log-likeihood function first, $$ begin{aligned} ell(mu|mathbf{x},mathbf{y})&=-log(mu^{n+m})-frac{1}{mu}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]\ &=-(n+m)log(mu)-frac{1}{mu}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]. end{aligned} $$ Taking the derivative with respect to $mu$, gives us $$ frac{operatorname{d}}{operatorname{d}mu}ell(mu|mathbf{x},mathbf{y})=-(n+m)frac{1}{mu}+frac{1}{mu^2}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright], $$ equate this to zero to obtain the stationary point, $$ begin{aligned} -(n+m)frac{1}{mu}+frac{1}{mu^2}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]&=0\ -(n+m)mu+left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]&=0\ mu&=frac{1}{n+m}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]. end{aligned} $$ To verify if this is the MLE, we take the second derivative test for the log-likelihood function, $$ frac{operatorname{d}^2}{operatorname{d}mu^2}ell(mu|mathbf{x},mathbf{y})=(n+m)frac{1}{mu^2}-frac{2}{mu^3}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]0}displaystylefrac{1}{mu^{n+m}}expleft{-frac{1}{mu}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]right}}{displaystylefrac{1}{bar{x}^n}frac{1}{bar{y}^m}exp[-(n+m)]}\ &=left(frac{1}{frac{1}{{(n+m)}^{n+m}}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]^{n+m}}timesright.\ &qquadleft.expleft{-frac{1}{frac{1}{n+m}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]right}right)bigg/\ &qquadqquadqquaddisplaystylefrac{1}{bar{x}^n}frac{1}{bar{y}^m}exp[-(n+m)] end{aligned} $$ $$ begin{aligned} &=frac{displaystylefrac{1}{displaystylefrac{1}{{(n+m)}^{n+m}}left[displaystylesum_{i=1}^nx_i+sum_{j=1}^m y_jright]^{n+m}}timesexp[-(n+m)]}{displaystylefrac{1}{bar{x}^n}frac{1}{bar{y}^m}exp[-(n+m)]}\[.3cm] &=frac{displaystyle bar{x}^n bar{y}^m}{displaystylefrac{1}{{(n+m)}^{n+m}}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]^{n+m}}. end{aligned} $$ And we say that $H_0$ is rejected if $lambda(mathbf{x},mathbf{y})leq c$.
      2. If we do some algebra on the LRT in part (a), we obtain the following: $$ begin{aligned} lambda(mathbf{x},mathbf{y})&=frac{displaystyle bar{x}^n bar{y}^m}{displaystylefrac{1}{{(n+m)}^{n+m}}left[sum_{i=1}^nx_i+sum_{j=1}^m y_jright]^{n+m}}\ &=frac{displaystylefrac{1}{n^n}left(sum_{i=1}^{n}x_iright)^{n}frac{1}{m^{m}}left(sum_{j=1}^{m}y_jright)^{m}}{displaystylefrac{1}{(n+m)^{n+m}}left[sum_{i=1}^{n}x_i+sum_{j=1}^{m}y_jright]^{n+m}}\ &=frac{displaystyle (n+m)^{n+m}left(sum_{i=1}^{n}x_iright)^{n}left(sum_{j=1}^{m}y_jright)^{m}}{displaystyle n^{n}m^{m}left[sum_{i=1}^{n}x_i+sum_{j=1}^{m}y_jright]^{n+m}}\ &=frac{(n+m)^{n+m}}{n^nm^{m}}left[frac{displaystyle sum_{j=1}^{m}y_j}{displaystylesum_{i=1}^{n}x_i+sum_{j=1}^{m}y_j}right]^{m}left[frac{displaystyle sum_{i=1}^{n}x_i}{displaystylesum_{i=1}^{n}x_i+sum_{j=1}^{m}y_j}right]^{n}\ &=frac{(n+m)^{n+m}}{n^nm^{m}}left[1-frac{displaystyle sum_{i=1}^{n}x_i}{displaystylesum_{i=1}^{n}x_i+sum_{j=1}^{m}y_j}right]^{m}left[frac{displaystyle sum_{i=1}^{n}x_i}{displaystylesum_{i=1}^{n}x_i+sum_{j=1}^{m}y_j}right]^{n}\ &=frac{(n+m)^{n+m}}{n^nm^{m}}left[1-Tright]^{m}left[Tright]^{n}. end{aligned} $$ Hence, the LRT can be based on the statistic $T$.
      3. The distribution of $sum X_i$ is obtain using the MGF (Moment Generating Function) technique, that is $$ begin{aligned} mathrm{M}_{Sigma X_i}(t)&=mathrm{E}exp[tSigma X_i]=mathrm{E}exp[tX_1 +cdots + tX_n]\ &=mathrm{E}exp[tX_1]timescdotstimesmathrm{E}exp[tX_n],quadtext{by independence.}\ &=frac{1}{1-theta t}timescdotstimesfrac{1}{1-theta t}\ &=left(frac{1}{1-theta t}right)^{n}=text{MGF of gamma}(n,theta). end{aligned} $$ Now, when $H_0$ is true then $sum X_i$ is gamma($m,theta$). For brevity, let $X=sum_{i=1}^{n} X_i$ and $Y=sum_{j=1}^{m}Y_j$. The joint distribution of $X$ and $Y$ is given below, $$ f_{XY}(x, y)=frac{1}{Gamma (n)theta^{n}}x^{n-1}exp[-x/theta]timesfrac{1}{Gamma (m)theta^{m}}y^{m-1}exp[-y/theta]. $$ Let $U=frac{X}{X+Y}$ and $V=X+Y$, then the support of $(X,Y)$ is $mathcal{A}=left{(x,y)in mathbb{R}^{+}times mathbb{R}^{+}right}$. Since the transformations $U$ and $V$ is one-to-one and onto, then $mathcal{B}=left{(u,v)in [0,1]times mathbb{R}^{+}right}$. Consider the following transformations $$ u=g_{1}(x,y)=frac{x}{x+y}quadtext{and}quad v=g_{2}(x,y)=x+y. $$ Then, begin{equation} label{eq:bvt1} u=frac{x}{x+y}Rightarrow x=frac{uy}{1-u} end{equation} and begin{equation} label{eq:bvt2} v=x+yRightarrow y = v-x. end{equation} Substitute Equation (ref{eq:bvt2}) to Equation (ref{eq:bvt1}), then $$ begin{aligned} x=frac{u(v-x)}{1-u}&Rightarrow x(1-u)=u(v-x)\ x-ux=uv-ux&Rightarrow x=uv=h_{1}(u,v). end{aligned} $$ Substitute $x$ above to Equation (ref{eq:bvt2}) to obtain, $$y=v(1-u)=h_2(u,v).$$ And the Jacobian matrix is, $$ mathbf{J}=bigg| begin{array}{cc} v&u\[.2cm] -v&1-u end{array} bigg|=v(1-u)+uv=v. $$ So that, $$ begin{aligned} f_{UV}(u,v)&=f_{XY}(h_1(u,v),h_2(u,v))lvert mathbf{J}rvert=f_{XY}(uv,v(1-u))lvert vrvert\ &=frac{1}{Gamma (n)theta^{n}}(uv)^{n-1}exp[-uv/theta]times\ &quad;frac{1}{Gamma (m)theta^{m}}(v(1-u))^{m-1}exp[-v(1-u)/theta]v\ &=frac{1}{Gamma (n)theta^{n}}(uv)^{n-1}exp[-uv/theta]times\ &quad;frac{1}{Gamma (m)theta^{m}}(v(1-u))^{m-1}exp[-v/theta]exp[uv/theta]v\ &=frac{1}{Gamma (n)theta^{n}}u^{n-1}v^{n-1}timesfrac{1}{Gamma (m)theta^{m}}v^{m-1}(1-u)^{m-1}exp[-v/theta]v\ &=frac{1}{Gamma (n)}underbrace{u^{n-1}(1-u)^{m-1}}_{text{Beta}(n,m)text{ kernel}}frac{1}{Gamma (m)theta^{m+n}}v^{m-1}v^{n-1}exp[-v/theta]v\ &=frac{Gamma(m)Gamma(m+n)}{Gamma(m)Gamma(m+n)}frac{u^{n-1}(1-u)^{m-1}}{Gamma (n)}times\ &quad;frac{1}{Gamma (m)theta^{m+n}}v^{m-1}v^{n}exp[-v/theta]\ &=underbrace{frac{Gamma(m+n)}{Gamma (n)Gamma(m)}u^{n-1}(1-u)^{m-1}}_{text{Beta}(n,m)}times\ &quad;underbrace{frac{1}{Gamma(m+n)theta^{m+n}}v^{m+n-1}exp[-v/theta]}_{text{Gamma}(m+n,theta)}. end{aligned} $$ So that the marginal density of $U=displaystylefrac{sum X_i}{sum X_i +sum Y_i}$ is Beta($n,m$).

      Reference

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

      Motivational quotes for the mind and soul.



      Lately, I’ve been feeling down and a bit sad. I’m anxious most of the time and I worry to much about the future. It’s one of those days when getting out of the bed is really really hard. But I have to drag myself up, I have to get up and get my shit together. I just have to… And that’s what I did this morning. 

      Yes, we have those moments where we feel unmotivated and uninspired, like we have the whole world in our shoulders. But you know what what helps? Aside from a good book and a cup of coffee, finding and reading motivational quotes can help a lot too! That’s what I do, I remind myself to get up and start moving forward.

      So here are some inspirational and motivational quotes that may help us go through our gloomy days:

        • ”A successful man is one who can lay a firm foundation with the bricks others have thrown at him.”   -David Brinkley
        • “If you want to live a happy life, ti it to a goal. Not to people or things.” -Albert Einstein
        • “Everyone has inside them a piece of good news. The good news is you don’t know how great you can be! How much you can love! What you can accomplish! And what your potential is.” -Anne Frank
        • “You have to fight through some bad days to earn the best days of your life.”  -Anonymous
        • “Every great dream begins with a dreamer. Always remember, you have within you the strength, the patience, and the passion to reach for the stars to change the world.” -Harriet Tubman
        • The best revenge is massive success.   –Frank Sinatra
        • “Success is not the key to happiness. Happiness is the key to success. If you love what you are doing, you will be successful.”   –Albert Schweitzer
        • “Character cannot be developed in ease and quiet. Only through experience of trial and suffering can the soul be strengthened, ambition inspired, and success achieved.”   -Helen Keller
        • You still have a lot of time to make yourself be what you want. There’s still lots of good in this world.   -Anonymous
        • “But I believe good things happen everyday. I believe good things happen even when bad things happen. And I believe on a happy day like today, we can still feel a little sad. And that’s life, isn’t it?”   -Elsewhere by Gabrielle Zevin

        I know i’m not alone on this one. Maybe somehow, you have experienced this too. I want you to know, you too are not alone. Just look around you. There are a lot of things to be grateful for.

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