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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_2with 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) 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

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_2with 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) 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!

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