Question

[1] The joint probability mass function of two discrete random variables A and B is Pab(a,b) = {ca?b, Sca²b, a= -2,2 and b = 1,2 otherwise Clearly stating your reasons, answer the following two (i) Are A and B are uncorrelated? (ii) Are A and B independent?

Answer 1

Firstly, we find the value of c. It can be found by noting that the sum of the values of the joint probability mass function summed over the support of A and B must equal to 1. Thus, we get:

$\begin{align*} & \ \ \ \ \sum_{a} \sum_{b} P_{A,B}(a,b) &&= 1 \\ &\Rightarrow P_{A,B}(-2,1) + P_{A,B}(-2,2) + P_{A,B}(2,1) + P_{A,B}(2,2) &&= 1 \\ &\Rightarrow c*(-2)^2*1 + c*(-2)^2*2 + c*2^2*1 + c*2^2*2 &&= 1 \\ &\Rightarrow 4c + 8c + 4c + 8c &&= 1 \\ &\Rightarrow 24c &&= 1 \\ &\Rightarrow c &&= \frac{1}{24} \end{align*}$

Thus, the joint probability mass function of A and B is given by:

4 PAB(-2, 1) = = 24 6 2 PAB(-2, 2) = = *(-2)2 * 1 24 1 *(-2)2 * 2 24 1 *22*1 24 1 *22*2 24 6 1 PAB(2, 1) = 24 4 24 8 6 PAB(2, 2) 24 6

Moreover, the marginal probability mass function of A is given by:

PA(-2) = PAB(-2, 1) +PAB(-2, 2) = + 6 O NON PA(2) = PA,B(2,1)+ PA,B(2, 2) = + 6

The marginal probability mass function of B is given by:

$\begin{align*} P_B(1) &= P_{A,B}(-2,1) + P_{A,B}(2,1) &&= \frac{1}{6} + \frac{1}{6} &&&= \frac{1}{3} \\ P_B(2) &= P_{A,B}(-2,2) + P_{A,B}(2,2) &&= \frac{2}{6} + \frac{2}{6} &&&= \frac{2}{3} \end{align*}$

(i)

To find whether A and B are uncorrelated, we find the covariance between A and B. For that, we first find E(AB), E(A) and E(B):

E(AB) = ab * PA,B(a, b) a b 1 1 = (-2) *1* PAB(-2,1)+(-2) * 2 * PA,B(-2,2) + 2* 1 * PAB(2,1) + 2 * 2 * PA,B(2, 2) 2 2 = -2 -4 - 2* = +4* 6 6 6 2 8 2 8 = 6 6 6 6 = + - + +

$\begin{align*} E(A) &= \sum_a a*P_A(a) \\ &= (-2)*P_A(-2) + 2*P_A(2) \\ &= -2*\frac{1}{2} + 2*\frac{1}{2} \\ &= -1 + 1 \\ &=0 \end{align*}$

$\begin{align*} E(B) &= \sum_b b*P_B(b) \\ &= 1*P_B(1) + 2*P_B(2) \\ &= 1*\frac{1}{3} + 2*\frac{2}{3} \\ &= \frac{1}{3} + \frac{4}{3} \\ &= \frac{5}{3} \end{align*}$

Now, we find the covariance between A and B:

$\begin{align*} Cov(A,B) &= E(AB) - E(A) E(B) \\ &= 0-0*\frac{5}{3} \\ &= 0-0 \\ &= 0 \end{align*}$

Since, the covariance between A and B is zero, A and B are uncorrelated. [ANSWER]

(ii)

For A and B to be independent it must satisfy $\begin{align*} P_{A,B}(a,b) = P_A(a)*P_B(b) \end{align*}$ for a = -2,2 and b = 1,2.

Now, we find:

$\begin{align*} P_A(-2)*P_B(1) &= \frac{1}{2}*\frac{1}{3} &&= \frac{1}{6} &&= P_{A,B}(-2,1) \\ P_A(-2)*P_B(2) &= \frac{1}{2}*\frac{2}{3} &&= \frac{2}{6} &&= P_{A,B}(-2,2) \\ P_A(2)*P_B(1) &= \frac{1}{2}*\frac{1}{3} &&= \frac{1}{6} &&= P_{A,B}(2,1) \\ P_A(2)*P_B(2) &= \frac{1}{2}*\frac{2}{3} &&= \frac{2}{6} &&= P_{A,B}(2,2) \end{align*}$

From above, we observe that $\begin{align*} P_{A,B}(a,b) = P_A(a)*P_B(b) \end{align*}$ for a = -2,2 and b = 1,2. Thus, A and B are independent. [ANSWER]

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