ISI MStat PSB 2014 Problem 4 | The Machine's Failure

This is a very simple sample problem from ISI MStat PSB 2014 Problem 4. It is based on order statistics, but generally due to one's ignorance towards order statistics, one misses the subtleties . Be Careful !

Problem- ISI MStat PSB 2014 Problem 4


Consider a machine with three components whose times to failure are independently distributed as exponential random variables with mean \(\lambda\). the machine continue to work as long as at least two components work. Find the expected time to failure of the machine.

Prerequisites


Exponential Distribution

Order statistics

Basic counting

Solution :

In the problem as it is said, let the 3 component part of the machine be A,B and C respectively, where \(X_A, X_B\) and \(X_C\) are the survival time of the respective parts. Now it is also told that \(X_A, X_B\) and \(X_C\) follows \(exponential(\lambda) \), and clearly these random variables are also i.id.

Now, here comes the trick ! It is told that the machine stops when two or all parts of the machine stop working. Here, we sometimes gets confused and start thinking combinatorially. But the we forget the basic counting of combinatorics lies in ordering ! Suppose we start ordering the life time of the individual components .i.e. among \(X_A, X_B\) and \(X_C\), there exists a ordering and say if we write it in order, we have \(X_{(1)} \le X_{(2)} \le X_{(3)} \).

Now observe that, after \(X_{(2)}\) units of time, the machine will stop !! (Are you sure ?? think it over ).

So, expected time till the machine stops , is just \(E(X_{(2)})\), but to find this we need to know the distribution of \(X_{(2)}\).

We have the pdf of \(X_{(2)}\) as, \(f_{(2)}(x)= \frac{3!}{(2-1)!(3-2)!} [P(X \le x)]^{2-1}[P(X>x)]^{3-2}f_X(x) \).

Where \(f_X(x)\) is the pdf of exponentional with mean \(\lambda\).

So, \(E(X(2))= \int^{\infty}_0 xf_{(2)}(x)dx \). which will turn out to be \(\frac{5\lambda}{6}\), which I leave on the readers to verify , hence concluding my solution.


Food For Thought

Now, suppose, you want install an alarm system, which will notify you some times before the machine wears our!! So, what do you think your strategy should be ? Given that you have a strategy, you now replace the weared out part of the machine within the time period between the alarm rings and the machine stops working, to continue uninterrupted working.What is the expected time within which you must act ?

Keep the machine running !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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ISI MStat PSB 2013 Problem 9 | Envelope Collector's Expenditure

This is a very simple and beautiful sample problem from ISI MStat PSB 2013 Problem 9. It is mainly based on geometric distribution and its expectation . Try it!

Problem- ISI MStat PSB 2013 Problem 9


Envelopes are on sale for Rs. 30 each. Each envelope contains exactly one coupon, which can be one of the one of four types with equal probability. Suppose you keep on buying envelopes and stop when you collect all the four type of coupons. What will be your expenditure ?

Prerequisites


Geometric Distribution

Expectation of geometric distribution

Basic counting

Solution :

This problem seems quite simple and it is simple, often one may argue that we can take a single random variable, which denotes the number of trials till the fourth success (or is it third !!), and calculate its expectation. But I differ here becaue I find its lot easier to work with sum of geometric random variables than a negative binomial. ( negative binomial is actually sum of finite geometrics !!)

So, here what we will do, is define 4 random variables, as \(X_i\) : # trials to get a type of coupon that is different from the all the \((i-1)\) types of coupons drawn earlier. \(i=1,2,3,4\).

Now since each type of coupon has an equal probability to come, that is probability of success is \(\frac{1}{4}\), here a common mistakes people commit is assuming all \(X_1,X_2,X_3,X_4\) are i.i.d Geometric(\(\frac{1}{4}\)), and this turns out to be a disaster !! So, be aware and observe keenly, that at the first draw, any of the four types will come, with probability 1, and there after we just need the rest of the 3 types to appear at least once. So, here \(X_1=1\) always and \(X_2 \sim Geo(\frac{3}{4})\)( becuase since in the first trial, surely any of the four types will come, our success would be getting any of the 3 types of envelopes from the all types making the success probability \(\frac{3}{4}\)) similarly , \(X_3 \sim Geo(\frac{1}{2})\) and \(X_4 \sim Geo(\frac{1}{4})\).

Now for the expectated expenditure at the given rate of Rs. 30 per envelope, expected expenditure is Rs.\(30 E(X_1+X_2+X_3+X_4)\)

Now, we know that \(E(X_2)=\frac{4}{3}\), \(E(X_3)=2\) and \(E(X_4)=4\) (why??)

So, \(E(X_1+X_2+X_3+X_4)=1+\frac{4}{3}+2+4=\frac{25}{3}\).

So, our required expectation is Rs. 250. Hence we are done !


Food For Thought

Suppose among the envelopes you collected each envelope has an unique 5 digit number, you picked an envelope randomly and what is the chance of the number that is going to show up is has its digits in a non-decreasing order ?

How does the chances may vary if you find a k digit number on the envelope ? think over it !!


ISI MStat PSB 2008 Problem 10
Outstanding Statistics Program with Applications

Outstanding Statistics Program with Applications

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Conditions and Chance | ISI MStat 2018 PSB Problem 5

This problem is a cute application of joint distribution and conditional probability. This is the problem 5 from ISI MStat 2018 PSB.

Problem

Suppose \(X_{1}\) and \(X_{2}\) are identically distributed random variables, not necessarily independent, taking values in \(\{1,2\}\). If \(\mathrm{E}\left(X_{1} X_{2}\right)= \frac{7}{3} \) and \(\mathrm{E}\left(X_{1}\right) = \frac{3}{2},\) obtain the joint distribution of \(\left(X_{1}, X_{2}\right)\).

Prerequisites

Solution

This problem is mainly about crunching the algebra of the conditions and get some good conditions for you to easily trail your path to the solution.

Usually, we go forward starting with the distribution of \(X_1\) and \(X_2\) to the distribution of (\(X_1, X_2\)). But, we will go backward from the distribution of (\(X_1, X_2\)) to \(X_1\), \(X_2\) and \(X_1X_2\)with the help of conditional probability.

conditional probability

Now, observe \(p_{21} = p_{12}\) because \(X_1\) and \(X_2\) are identically distributed.

Let's calculate the following:

\(P(X_1 = 1 )= p_{11} + p_{12} = P(X_2 = 1)\)

\(P(X_1 = 2) = p_{12} + p_{22} = P(X_2 = 2)\)

\(E(X_1) = p_{11} + 3p_{12} + 2p_{22} = \frac{3}{2}\)

Now, \(X_1X_2\) can take values {\(1, 2, 4\)}.

\(X_1 = 1, X_2 = 1 \iff X_1X_2 = 1\) \( \Rightarrow P(X_1X_2 = 2) = p_{11}\).

\(X_1 = 2, X_2 = 2 \iff X_1X_2 = 4\) \( \Rightarrow P(X_1X_2 = 4) = p_{22}\).

\(X_1 = 1, X_2 = 1\) or \(X_1 = 2, X_2 = 1 \iff X_1X_2 = 2\) \( \Rightarrow P(X_1X_2 = 2) = 2p_{12}\).

\(E(X_1X_2) = p_{11} + 4p_{12} + 4p_{22} = \frac{7}{3}\).

Now, we need another condition, do you see that ?

\(p_{11} + 2p_{12} + p_{44} = 1\).

Now, you can solve it easily to get the solutions \( p_{11} = \frac{1}{3}, p_{12} = \frac{1}{6}, p_{22} =\frac{1}{3} \).

Food for Thought

Now, what do you think, how many expectation values will be required if \(X_1\) and \(X_2\) takes values in {1, 2, 3}?

What if \(X_1\) and \(X_2\) takes values in {\(1, 2, 3, 4, ..., n\)}?

What if there are \(X_1, X_2, ...., X_n\) taking values in {\(1, 2, 3, 4, ..., m\)}?

This is just another beautiful counting problem.

Enjoy and Stay Tuned!