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

Probability Theory
Joint Distribution
The expectation of Functions of Random Variables
Conditional Probability

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.

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

Let's calculate the following:

$P(X_1 = 1)$ and $P(X_1 = 2)$ and $E(X_1)$
$P(X_2 = 1)$ and $P(X_2 = 2)$
$E(X_1X_2)$

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

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

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

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!

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

Problem

Prerequisites

Probability Theory
Joint Distribution
The expectation of Functions of Random Variables
Conditional Probability

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.

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

Let's calculate the following:

$P(X_1 = 1)$ and $P(X_1 = 2)$ and $E(X_1)$
$P(X_2 = 1)$ and $P(X_2 = 2)$
$E(X_1X_2)$

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

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

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

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!

Conditions and Chance | ISI MStat 2018 PSB Problem 5

Problem

Prerequisites

Solution

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

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

$p_{11} + 2p_{12} + p_{44} = 1$ .

Food for Thought

Related

Problem

Prerequisites

Solution

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

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

$p_{11} + 2p_{12} + p_{44} = 1$ .

Food for Thought

Related

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