Q2Advanced Engineering Mathematics

Question

The joint probability mass function of (X,Y) is given by P_{X Y} (x, y) = {λ x^{2} y 0 i = 1, 2; j = 1, 2, 3 otherwise i) Find λ ii) Find the marginal probability mass function of x and y.

Answer 1

We are given the joint probability mass function (PMF) of two discrete random variables $X$ and $Y$ defined as $P_{X Y} (x, y) = λ x^{2} y$ for the specific domain $x \in {1, 2}$ and $y \in {1, 2, 3}$ . For all other values, the probability is zero.

By the fundamental axiom of probability, the sum of a joint probability mass function over the entire defined sample space must be exactly equal to 1. Therefore, we evaluate the double summation over all possible values of $x$ and $y$ :

x = 1 \sum 2 y = 1 \sum 3 P_{X Y} (x, y) = 1

Substituting the given function into the equation:

x = 1 \sum 2 y = 1 \sum 3 λ x^{2} y = 1

We can factor out $λ$ and separate the summations because the terms for $x$ and $y$ are separable:

λ (x = 1 \sum 2 x^{2}) (y = 1 \sum 3 y) = 1

Now, calculate the value of each summation independently:

x = 1 \sum 2 x^{2} = 1^{2} + 2^{2} = 1 + 4 = 5

y = 1 \sum 3 y = 1 + 2 + 3 = 6

Substituting these values back into our main equation:

λ (5) (6) = 1 ⟹ 30 λ = 1 ⟹ λ = \frac{1}{30}

The marginal probability mass function describes the probability distribution of one variable while completely ignoring the values of the other variable. It is found by summing the joint probability over all possible values of the other variable.

Marginal PMF of X, $P_{X} (x)$ : We sum the joint PMF over all possible values of $Y$ (which are 1, 2, 3):

P_{X} (x) = y = 1 \sum 3 P_{X Y} (x, y) = y = 1 \sum 3 \frac{1}{30} x^{2} y = \frac{x ^{2}}{30} y = 1 \sum 3 y = \frac{x ^{2}}{30} (1 + 2 + 3) = \frac{6 x ^{2}}{30} = \frac{x ^{2}}{5}

This is valid for $x = 1, 2$ .

Marginal PMF of Y, $P_{Y} (y)$ : We sum the joint PMF over all possible values of $X$ (which are 1, 2):

P_{Y} (y) = x = 1 \sum 2 P_{X Y} (x, y) = x = 1 \sum 2 \frac{1}{30} x^{2} y = \frac{y}{30} x = 1 \sum 2 x^{2} = \frac{y}{30} (1^{2} + 2^{2}) = \frac{5 y}{30} = \frac{y}{6}

This is valid for $y = 1, 2, 3$ .

Q2Advanced Engineering Mathematics

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