ISI MStat PSB 2005 Problem 5 | Uniformity of Uniform

This is a simple and elegant sample problem from ISI MStat PSB 2005 Problem 5. It's based the mixture of Discrete and Continuous Uniform Distribution, the simplicity in the problem actually fools us, and we miss subtle happenings. Be careful while thinking !

Problem- ISI MStat PSB 2005 Problem 5

Suppose $X$ and $U$ are independent random variables with

$P(X=k)=\frac{1}{N+1}$ , $k=0,1,2,......,N$ ,

and $U$ having a uniform distribution on [0,1]. Let $Y=X+U$ .

(a) For $y \in \mathbb{R}$ , find $P(Y \le y)$ .

(b) Find the correlation coefficient between $X$ and $Y$ .

Prerequisites

Uniform Distribution

Law of Total Probability

Conditional Distribution

Solution :

This ptroblem is quite straight forward enough, and we do what we are told to.

Here, we need to find the Cdf of $Y$ , where , $Y=X+U$ , and $X$ and $Y$ are defined as above.

So, $P(Y\le y)= P(X+U \le y)=\sum_{k=0}^N P(U \le y-k|X=k)P(X=k) = \frac{1}{N+1} \sum_{i=1}^NP(U \le y-k)$ , [ since $U$ and $X$ are indepemdent ],

Now, here is where we get fooled often by the simplicity of the problem. The beauty is to observe in the above expression, if $y-k <0$ then $P(U\le y-k)=0$ , and if $y-k>1$ then $P(U\le y-k)=1$ , ( why??)

So, for $k* \le y \le k*+1$ the $P(U \le y-k*)=y-k*$ , so when $k* \le y \le k*+1$ , $P(U\le y-k)=0$ for $k>k*$ , and $P(U \le y-k)=1$ for $k<k*$ .

So, the required Cdf will depend on the interval , y belongs to, for the above illustrated case, i.e. $k* \le y\le k*+1$ there will be $k*$ number of 1's, and $N-k*-1$ number of 0's in the above summation, derived, so, $P(Y\le y)$ reduces to,

$P(Y\le y)= \frac{1}{N+1} ( k*+y-k*)=\frac{y}{N+1}$ , $0<y<N+1$ , [ since here $k*$ may vary from 0,1,...., N, hence union of all the nested sub-intervals give the mentioned interval]

Hence the required Cdf. But can you find this Cdf argumentatively, without this algebraic labor ?? What distribution is it ?? Does the Uniform retains its Uniformity ?

I leave the part (b) as an exercise, its quite trivial.

Food For Thought

How to deal with some random triangles ??

Suppose, You are given a stick of length $N+1$ units, Now you are to break the stick into 3 pieces, and the breaking points are chosen randomly , What is the chance of constructing a triangle with those 3 broken part as the sides of the constructed triangle ??

Now if you first break the stick into two pieces randomly, and further break the longer piece again into two parts (randomly), How the chance of making a triangle changes ?? Do the chance increases what do you think ??

Lastly, does the length of the stick matters at all , Give it a thought !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

This is a simple and elegant sample problem from ISI MStat PSB 2005 Problem 5. It's based the mixture of Discrete and Continuous Uniform Distribution, the simplicity in the problem actually fools us, and we miss subtle happenings. Be careful while thinking !

Problem- ISI MStat PSB 2005 Problem 5

Suppose $X$ and $U$ are independent random variables with

$P(X=k)=\frac{1}{N+1}$ , $k=0,1,2,......,N$ ,

and $U$ having a uniform distribution on [0,1]. Let $Y=X+U$ .

(a) For $y \in \mathbb{R}$ , find $P(Y \le y)$ .

(b) Find the correlation coefficient between $X$ and $Y$ .

Prerequisites

Uniform Distribution

Law of Total Probability

Conditional Distribution

Solution :

This ptroblem is quite straight forward enough, and we do what we are told to.

Here, we need to find the Cdf of $Y$ , where , $Y=X+U$ , and $X$ and $Y$ are defined as above.

So, $P(Y\le y)= P(X+U \le y)=\sum_{k=0}^N P(U \le y-k|X=k)P(X=k) = \frac{1}{N+1} \sum_{i=1}^NP(U \le y-k)$ , [ since $U$ and $X$ are indepemdent ],

Now, here is where we get fooled often by the simplicity of the problem. The beauty is to observe in the above expression, if $y-k <0$ then $P(U\le y-k)=0$ , and if $y-k>1$ then $P(U\le y-k)=1$ , ( why??)

So, for $k* \le y \le k*+1$ the $P(U \le y-k*)=y-k*$ , so when $k* \le y \le k*+1$ , $P(U\le y-k)=0$ for $k>k*$ , and $P(U \le y-k)=1$ for $k<k*$ .

So, the required Cdf will depend on the interval , y belongs to, for the above illustrated case, i.e. $k* \le y\le k*+1$ there will be $k*$ number of 1's, and $N-k*-1$ number of 0's in the above summation, derived, so, $P(Y\le y)$ reduces to,

$P(Y\le y)= \frac{1}{N+1} ( k*+y-k*)=\frac{y}{N+1}$ , $0<y<N+1$ , [ since here $k*$ may vary from 0,1,...., N, hence union of all the nested sub-intervals give the mentioned interval]

Hence the required Cdf. But can you find this Cdf argumentatively, without this algebraic labor ?? What distribution is it ?? Does the Uniform retains its Uniformity ?

I leave the part (b) as an exercise, its quite trivial.

Food For Thought

How to deal with some random triangles ??

Suppose, You are given a stick of length $N+1$ units, Now you are to break the stick into 3 pieces, and the breaking points are chosen randomly , What is the chance of constructing a triangle with those 3 broken part as the sides of the constructed triangle ??

Now if you first break the stick into two pieces randomly, and further break the longer piece again into two parts (randomly), How the chance of making a triangle changes ?? Do the chance increases what do you think ??

Lastly, does the length of the stick matters at all , Give it a thought !!

Similar Problems and Solutions

ISI MStat PSB 2008 Problem 10 — Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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