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# ISI MStat PSB 2010 Problem 10 | Uniform Modified This is a very elegant sample problem from ISI MStat PSB 2010 Problem 10. It's mostly based on propertes of uniform, and its behaviour when modified . Try it!

## Problem- ISI MStat PSB 2010 Problem 10

Let $X$ be a random variable uniformly distributed over $(0,2\theta )$, $\theta>0$, and $Y=max(X,2\theta -X)$.

(a) Find $\mu =E(Y)$.

(b) Let $X_1,X_2,.....X_n$ be a random sample from the above distribution with unknown $\theta$. Find two distinct unbiased estimators of $\mu$, as defined in (a), based on the entire sample.

### Prerequisites

Uniform Distribution

Law of Total Expectation

Unbiased Estimators

## Solution :

Well, this is a very straight forward problem, where we just need to be aware the way $Y$ is defined.

As, we need $E(Y)$ and by definition of $Y$ , we clearly see that $Y$ is dependent in $X$ where $X \sim Unif( 0, 2\theta)$.

So, using Law of Total Expectation,

$E(Y)= E(X|X>2\theta-X)P(X>2\theta-X)+E(2\theta-X|X \le 2\theta-X)P(X \le 2\theta-X). Observe that, \(P(X \le \theta)=\frac{1}{2}$, why ??

Also, conditional pdf of $X|X>\theta$ is,

$f_{X|X>\theta}(x)=\frac{f_X(x)}{P(X>\theta)}==\frac{1}{\theta} \& \theta< x \le 2\theta$. [where $f_X$ is the pdf of $X$].

the other conditional pdf is also same due to symmetry.(Verify!!).

So, $E(Y)=E(X|X\sim Unif(\theta,2\theta))\frac{1}{2}+E(X|X\sim Unif(0,\theta))\\frac{1}{2}=\frac{1}{2}(\frac{3\theta}{2}+2\theta-\frac{\theta}{2})=\frac{3\theta}{2}$.

hence, $\mu=\frac{3\theta}{2}$.

Now, for the next part, one trivial unbiased estimator of $\theta$ is $T_n=\frac{1}{n}\sum_{i=1}^n X_i$ (based on the given sample). So,

$\frac{3T_n}{2}=\frac{3}{2n}\sum_{i=1}^n X_i$ is an obvious unbiased estimator of $\mu$.

For another we need to change our way of looking on conventional way and look for the order statistics, since we know that $X_{(n)}$ is sufficient for $\theta$.(Don't Know ?? Look for Factorization Theorem .)

So, verify that $E(X_{(n)})=\frac{2n}{n+1}\theta$.

Hence, $\frac{n+1}{2n}X_{(n)}$ is another unbiased estimator of $theta$. So, $\frac{3(n+1)}{4n}X_{(n)}$ is also another unbiased estimator of $\mu$ a defined in (a).

Hence the solution concludes.

## Food For Thought

Let us think about some unpopular but very beautiful relationship between discrete random variables besides the Universality of uniform. Let $X$be a discrete random variable with cdf $F_X(x)$ and define the random variable $Y=F_X(x)$.

Can you verify that, $Y$ is stochastically greater that a uniform(0,1) random variable $U$. i.e.

$P(Y>y) \ge P(U>y)=1-y$ for all $y$, $0<y<1$,

$P(Y>y) > P(U>y) =1-y$, for some $y$, $0<y<1$.

Hint: Draw a typical picture of a discrete cdf, and observe the jump points ! you may jump to the solution!! Think it over.

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This is a very elegant sample problem from ISI MStat PSB 2010 Problem 10. It's mostly based on propertes of uniform, and its behaviour when modified . Try it!

## Problem- ISI MStat PSB 2010 Problem 10

Let $X$ be a random variable uniformly distributed over $(0,2\theta )$, $\theta>0$, and $Y=max(X,2\theta -X)$.

(a) Find $\mu =E(Y)$.

(b) Let $X_1,X_2,.....X_n$ be a random sample from the above distribution with unknown $\theta$. Find two distinct unbiased estimators of $\mu$, as defined in (a), based on the entire sample.

### Prerequisites

Uniform Distribution

Law of Total Expectation

Unbiased Estimators

## Solution :

Well, this is a very straight forward problem, where we just need to be aware the way $Y$ is defined.

As, we need $E(Y)$ and by definition of $Y$ , we clearly see that $Y$ is dependent in $X$ where $X \sim Unif( 0, 2\theta)$.

So, using Law of Total Expectation,

$E(Y)= E(X|X>2\theta-X)P(X>2\theta-X)+E(2\theta-X|X \le 2\theta-X)P(X \le 2\theta-X). Observe that, \(P(X \le \theta)=\frac{1}{2}$, why ??

Also, conditional pdf of $X|X>\theta$ is,

$f_{X|X>\theta}(x)=\frac{f_X(x)}{P(X>\theta)}==\frac{1}{\theta} \& \theta< x \le 2\theta$. [where $f_X$ is the pdf of $X$].

the other conditional pdf is also same due to symmetry.(Verify!!).

So, $E(Y)=E(X|X\sim Unif(\theta,2\theta))\frac{1}{2}+E(X|X\sim Unif(0,\theta))\\frac{1}{2}=\frac{1}{2}(\frac{3\theta}{2}+2\theta-\frac{\theta}{2})=\frac{3\theta}{2}$.

hence, $\mu=\frac{3\theta}{2}$.

Now, for the next part, one trivial unbiased estimator of $\theta$ is $T_n=\frac{1}{n}\sum_{i=1}^n X_i$ (based on the given sample). So,

$\frac{3T_n}{2}=\frac{3}{2n}\sum_{i=1}^n X_i$ is an obvious unbiased estimator of $\mu$.

For another we need to change our way of looking on conventional way and look for the order statistics, since we know that $X_{(n)}$ is sufficient for $\theta$.(Don't Know ?? Look for Factorization Theorem .)

So, verify that $E(X_{(n)})=\frac{2n}{n+1}\theta$.

Hence, $\frac{n+1}{2n}X_{(n)}$ is another unbiased estimator of $theta$. So, $\frac{3(n+1)}{4n}X_{(n)}$ is also another unbiased estimator of $\mu$ a defined in (a).

Hence the solution concludes.

## Food For Thought

Let us think about some unpopular but very beautiful relationship between discrete random variables besides the Universality of uniform. Let $X$be a discrete random variable with cdf $F_X(x)$ and define the random variable $Y=F_X(x)$.

Can you verify that, $Y$ is stochastically greater that a uniform(0,1) random variable $U$. i.e.

$P(Y>y) \ge P(U>y)=1-y$ for all $y$, $0<y<1$,

$P(Y>y) > P(U>y) =1-y$, for some $y$, $0<y<1$.

Hint: Draw a typical picture of a discrete cdf, and observe the jump points ! you may jump to the solution!! Think it over.

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