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 !!

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Bayes comes to rescue | ISI MStat PSB 2007 Problem 7

This is a very beautiful sample problem from ISI MStat PSB 2007 Problem 7. It's a very simple problem, which very much rely on conditioning and if you don't take it seriously, you will make thing complicated. Fun to think, go for it !!

Problem- ISI MStat PSB 2007 Problem 7

Let \(X\) and \(Y\) be i.i.d. exponentially distributed random variables with mean \(\lambda >0 \). Define \(Z\) by :

\( Z = \begin{cases} 1 & if X <Y \\ 0 & otherwise \end{cases} \)

Find the conditional mean, \( E(X|Z=1) \).

Prerequisites

Conditional Distribution

Bayes Theorem

Exponential Distribution

Solution :

This is a very simple but very elegant problem to describe an unique and efficient technique to solve a class of problems, which may seem analytically difficult.

Here, for \(X\), \(Y\) and \(Z\) as defined in the question, lets find out what we need first of all.

Sometimes, breaking a seemingly complex problem into some simpler sub-problems, makes our way towards the final solution easier. In this problem, the possible simpler sub problems, which I think would help us is, "Whats the value of \(P(X<Y)\) (or similarly \(P(Z=1)\)) ?? ", "what is pdf of \(X|X<Y\)( or equivalently \(X|Z=1\)) ?" and finally " what is the conditional mean \(E(X|Z=1)\) ??". We will attain these questions one by one.

for the very first question, "Whats the value of \(P(X<Y)\) (or similarly \(P(Z=1)\)) ?? ", well the answer to this question is relatively simple, and I leave it as an exercise !! the probability value which one will find if done correctly is \( \frac{1}{2}\). Verify it, then only move forward!!

The 2nd question is the most vital and beautiful part of the problem, we generally, do this kind of problems using the general definition of conditional probability, which you can obviously try, but will face some difficulties, which can be easily ignored by using the continuous form of Bayes' rule, which we are not often encouraged to use !! I don't really know why, though !

Let, find the conditional Cdf of \(X|Z=1\),

\( P(X \le x|Z=1) = \int^x_0 f_{X|X<Y}(x) dx, ........... x>0 \)

where \( f_{X|X<Y}(x)\) is the conditional pdf, which we are interested in, So now we can use Bayes rule on \(f_{X|X<Y}(x)\), we have,

\( f_{X|X<Y}(x)\)=\( \frac{P(Z=1|X=x)f_X(x)}{P(Z=1)}\)=\(\frac{P(Y>x)f_X(x)}{P(X<Y)}\)=\(\frac{\frac{e^{-\frac{x}{\lambda}}e^{- \frac{x}{\lambda}}}{\lambda}}{\frac{1}{2}}\)=\(\frac{2}{\lambda}e^{-\frac{2x}{\lambda}}\)

plugging this in the form of cdf we can easily verify, that \(X|Z=1 \sim expo(\frac{\lambda}{2}) \). (We can't say this directly from pdf because, pdfs are not unique, Can you give such an example ? think about it !)

So, now as we successfully answered the first 2 questions its easy to, answer the last and the final one, as \(X|Z=1 \sim expo(\frac{\lambda}{2}) \), its mean .i.e.

\(E(X|Z=1)=\frac{\lambda}{2}.\)

Hence the solution concludes.

Food For Thought

Lets, provide an interesting problem before concluding,

There, are K+1 machineas in a shop, all engaged in the mass production of an item. the \(i\)th machine produces defectives with probability of \(\frac{i}{k}\), i=0,1,2,.....,k.A machine is selected at random and then the items produced are repeatedly sampled. If the first n products are all defectives, then show that the conditional probability that (n+1)th sampled product will also be defective is approximately, equal to \( \frac{(n+1)}{(n+2)}\) when k is large.

Can you show it? Give it a try !!

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ISI MStat PSB 2008 Problem 8 | Bivariate Normal Distribution

This is a very beautiful sample problem from ISI MStat PSB 2008 Problem 8. It's a very simple problem, based on bivariate normal distribution, which again teaches us that observing the right thing makes a seemingly laborious problem beautiful . Fun to think, go for it !!

Problem- ISI MStat PSB 2008 Problem 8

Let \( \vec{Y} = (Y_1,Y_2)' \) have the bivariate normal distribution, \( N_2( \vec{0}, \sum ) \),

where, \(\sum\)= \begin{pmatrix} \sigma_1^2 & \rho\sigma_1\sigma_2 \\ \rho\sigma_2\sigma_1 & \sigma^2 \end{pmatrix} ;

Obtain the mean ad variance of \( U= \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2} \) .

Prerequisites

Bivariate Normal

Conditonal Distribution of Normal

Chi-Squared Distribution

Solution :

This is a very simple and cute problem, all the labour reduces once you see what to need to see !

Remember , the pdf of \(N_2( \vec{0}, \sum)\) ?

Isn't \( \vec{Y}\sum^{-1}\vec{Y}\) is the exponent of e, in the pdf of bivariate normal ?

So, we can say \(\vec{Y}\sum^{-1}\vec{Y} \sim {\chi_2}^2 \) . Can We ?? verify it !!

Also, clearly \( \frac{Y_1^2}{\sigma^2} \sim {\chi_1}^2 \) ; since \(Y_1\) follows univariate normal.

So, expectation is easy to find accumulating the above deductions, I'm leaving it as an exercise .

Calculating the variance may be a laborious job at first, but now lets imagine the pdf of the conditional distribution of \( Y_2 |Y_1=y_1 \) , what is the exponent of e in this pdf ?? \( U = \vec{Y'} {\sum}^{-1}\vec{Y} - \frac{Y_1^2}{\sigma^2} \) , right !!

and also , \( U \sim \chi_1^2 \) . Now doing the last piece of subtle deduction, and claiming that \(U\) and \( \frac{Y_1^2}{\sigma^2} \) are independently distributed . Can you argue why ?? go ahead . So, \( U+ \frac{Y_1^2}{\sigma^2} \sim \chi_2^2 \).

So, \( Var( U + \frac{Y_1^2}{\sigma^2})= Var( U) + Var( \frac{Y_1^2}{\sigma^2}) \)

\( \Rightarrow Var(U)= 4-2=2 \) , [ since, Variance of a R.V following \(\chi_n^2\) is \(2n\).]

Hence the solution concludes.

Food For Thought

Before leaving, lets broaden our mind and deal with Multivariate Normal !

Let, \(\vec{X}\) be a 1x4 random vector, such that \( \vec{X} \sim N_4(\vec{\mu}, \sum ) \), \(\sum\) is positive definite matrix, then can you show that,

\( P( f_{\vec{X}}(\vec{x}) \ge c) = \begin{cases} 0 & c \ge \frac{1}{4\pi^2\sqrt{|\sum|}} \\ 1-(\frac{k+2}{2})e^{-\frac{k}{2}} & c < \frac{1}{4\pi^2\sqrt{|\sum|}} \end{cases}\)

Where, \( k=-2ln(4\pi^2c \sqrt{|\sum|}) \).

Keep you thoughts alive !!

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