Question:
If X and Y are two random variables with the joint probability density function
\(f(x,y)=\left\{ \begin{array}{l} \frac{2}{3}(x+2y);\ \text{for}\ 0\lt x, y\lt 1 \\ 0;\ \ \ \ \ \ \ \ \ \ \text{otherwise} \end{array} \right.\)
then \(E[X|Y=\frac{1}{2}]\) will be
If X and Y are two random variables with the joint probability density function
\(f(x,y)=\left\{ \begin{array}{l} \frac{2}{3}(x+2y);\ \text{for}\ 0\lt x, y\lt 1 \\ 0;\ \ \ \ \ \ \ \ \ \ \text{otherwise} \end{array} \right.\)
then \(E[X|Y=\frac{1}{2}]\) will be
\(f(x,y)=\left\{ \begin{array}{l} \frac{2}{3}(x+2y);\ \text{for}\ 0\lt x, y\lt 1 \\ 0;\ \ \ \ \ \ \ \ \ \ \text{otherwise} \end{array} \right.\)
then \(E[X|Y=\frac{1}{2}]\) will be
Updated On: Aug 21, 2025
- \(\frac{5}{9}\)
- \(\frac{4}{9}\)
- \(\frac{1}{3}\)
- \(\frac{2}{3}\)
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The Correct Option is A
Solution and Explanation
Step 1: Marginal PDF of \(Y\)
\[ f_Y(y) = \int_0^1 f(x,y)\,dx = \int_0^1 \tfrac{2}{3}(x+2y)\,dx \] Compute the integral: \[ f_Y(y) = \tfrac{2}{3}\left[ \frac{x^2}{2} + 2yx \right]_0^1 = \tfrac{2}{3}\left( \tfrac{1}{2} + 2y \right) = \tfrac{1+4y}{3}, \quad 0
Step 2: Conditional PDF of \(X\) given \(Y=y\)
\[ f_{X|Y}(x|y) = \frac{f(x,y)}{f_Y(y)} = \frac{\tfrac{2}{3}(x+2y)}{\tfrac{1+4y}{3}} = \frac{2(x+2y)}{1+4y}, \quad 0
Step 3: Conditional Expectation
By definition, \[ E[X|Y=y] = \int_0^1 x \, f_{X|Y}(x|y)\, dx. \] Substituting, \[ E[X|Y=y] = \int_0^1 x \cdot \frac{2(x+2y)}{1+4y}\, dx = \frac{2}{1+4y} \int_0^1 (x^2 + 2yx)\, dx. \] Evaluate the integrals: \[ \int_0^1 x^2 dx = \frac{1}{3}, \quad \int_0^1 2yx\,dx = y. \] So, \[ E[X|Y=y] = \frac{2}{1+4y}\left( \tfrac{1}{3} + y \right). \]
Step 4: Substitute \(y=\tfrac{1}{2}\)
\[ E[X|Y=\tfrac{1}{2}] = \frac{2}{1+4\cdot \tfrac{1}{2}}\left( \tfrac{1}{3} + \tfrac{1}{2} \right) = \frac{2}{3}\left( \tfrac{5}{6} \right). \] \[ E[X|Y=\tfrac{1}{2}] = \tfrac{5}{9}. \]
Final Answer:
\[ \boxed{\tfrac{5}{9}} \]
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