Questions: The marginal distributions of two random variables X and Y, together with some values of their joint distribution, are given in the table below. Fill in the table in such a way that X and Y are independent random variables. Y=-1 Y=1 Y=2 ----------------------- X=-3 0.08 0.01 0.01 0.10 X=0 0.06 0.60 X=3 0.03 0.30 0.80 0.10 0.10 1.00

Solution Steps

To ensure that the random variables \(X\) and \(Y\) are independent, the joint probability \(P(X=x, Y=y)\) should equal the product of the marginal probabilities \(P(X=x)\) and \(P(Y=y)\) for all combinations of \(x\) and \(y\). We will use this property to fill in the missing values in the table. Specifically, for each missing value, we will calculate it as the product of the corresponding marginal probabilities.

To fill in the missing values in the joint distribution table, we use the independence property of the random variables \(X\) and \(Y\). The joint probability \(P(X=x, Y=y)\) can be calculated as:

\[ P(X=0, Y=-1) = P(X=0) \cdot P(Y=-1) = 0.6 \cdot 0.8 = 0.48 \]

\[ P(X=0, Y=2) = P(X=0) \cdot P(Y=2) = 0.6 \cdot 0.1 = 0.06 \]

\[ P(X=3, Y=-1) = P(X=3) \cdot P(Y=-1) = 0.3 \cdot 0.8 = 0.24 \]

\[ P(X=3, Y=2) = P(X=3) \cdot P(Y=2) = 0.3 \cdot 0.1 = 0.03 \]

Using the calculated values, we can now fill in the joint distribution table:

\[ \begin{array}{|c|c|c|c|c|} \cline{2-5} & Y=-1 & Y=1 & Y=2 & \\ \hline X=-3 & 0.08 & 0.01 & 0.01 & 0.10 \\ \hline X=0 & 0.48 & 0.06 & 0.06 & 0.60 \\ \hline X=3 & 0.24 & 0.03 & 0.03 & 0.30 \\ \hline & 0.80 & 0.10 & 0.10 & 1.00 \\ \hline \end{array} \]

Final Answer