Question

If \((X,Y)\) is a random vector with the joint probability mass function

Hint:

When finding conditional variance, first identify the conditional distribution. Adding a constant does not change variance.

Answer 1

Correct Answer: 0.75

Step 1: Write the conditional distribution for \(Y|X=2\).
For \(X=2\), we have
\[ y=2,3,4,\ldots \] The joint pmf becomes proportional to
\[ 3^{2(2)-y-1}=3^{3-y} \] Let
\[ K=Y-2 \] Then
\[ K=0,1,2,\ldots \] and
\[ y=2+k \]

Step 2: Express the probability in terms of \(K\).
\[ 3^{3-y}=3^{3-(2+k)} \] \[ =3^{1-k} \] Thus, the conditional distribution of \(K\) is proportional to
\[ 3^{1-k},\qquad k=0,1,2,\ldots \]
Now,
\[ \sum_{k=0}^{\infty}3^{1-k} = 3\sum_{k=0}^{\infty}\left(\frac13\right)^k \] \[ = 3\cdot \frac{1}{1-\frac13} \] \[ = 3\cdot \frac32 \] \[ = \frac92 \] Therefore,
\[ P(K=k|X=2) = \frac{3^{1-k}}{\frac92} \] \[ = \frac{2}{3}\left(\frac13\right)^k, \qquad k=0,1,2,\ldots \]

Step 3: Identify the distribution.
Thus, \(K\) follows a geometric distribution on \(0,1,2,\ldots\) with
\[ p=\frac23 \] and
\[ q=1-p=\frac13 \] For this geometric distribution,
\[ Var(K)=\frac{q}{p^2} \] So,
\[ Var(K)=\frac{\frac13}{\left(\frac23\right)^2} \] \[ = \frac{\frac13}{\frac49} \] \[ = \frac34 \] \[ =0.75 \]

Step 4: Use \(Y=K+2\).
Since
\[ Y=K+2, \] adding a constant does not change variance. Therefore,
\[ Var(Y|X=2)=Var(K) \] \[ =0.75 \]

Step 5: Final conclusion.
Hence,
\[ \boxed{0.75} \]