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Unit 04 /
Problem 1:


Conditional expectation and variance
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Easy


(Stacking Blocks)

In $\mathbb{R}^2$, define square region A with corners at $\{ (0,1), (1,1),(1,2), (0,2) \}$ and square region B with corners at $\{ (0.5,0),(1.5,0),(1.5,1),(0.5,1) \}$. Suppose that random variables X and Y have joint density given by,

$$f(x,y) = \begin{cases} 0.5, &(x,y) \in A \cup B \\
0, &otherwise \end{cases} $$

  1. Compute the conditional expectation function of Y given X.
  2. Compute the conditional expectation function of X given Y.
Solution:
  1. To compute the conditional expectation functions, we need to find the conditional probability density functions (PDFs) first. Once we have those, we can calculate the conditional expectations.

The conditional PDF of Y given X is given by:$$f_{Y∣X}(y∣x)=\frac{f(x,y)}{f_X(x)}$$​

Let's compute $f_X(x)$ first:

For $0 \leq x \leq 0.5$:

$$f_X(x) = \int_{1}^{2} 0.5 dy = 0.5$$

For $0.5 \leq x \leq 1$:

$$f_X(x) = \int_{0}^{2} 0.5 dy = 1$$

For $1 \leq x \leq 1.5$:

$$f_X(x) = \int_{0}^{1} 0.5 dy = 0.5$$

So the $f_X(x)$ is given by:

$$f_X(x) = \begin{cases} 0.5, & 0 \leq x \leq 0.5 \\
1, & 0.5 \leq x \leq 1 \\
0.5, & 1 \leq x \leq 1.5\end{cases}$$

And the $f_{Y∣X}(y∣x)$ is given by:

$$f_{Y∣X}(y∣x) = \begin{cases} 1, & 0 \leq x \leq 0.5 \\
0.5, & 0.5 \leq x \leq 1 \\
1, & 1 \leq x \leq 1.5\end{cases}$$

Now, let's compute the conditional expectation of Y given X:

For $0 \leq x \leq 0.5$:

$$E(Y|X) = \int_{1}^{2} y dy = 1.5$$

For $0.5 \leq x \leq 1$:

$$E(Y|X) = \int_{0}^{2} 0.5y dy = 1$$

For $1 \leq x \leq 1.5$ is:

$$E(Y|X) = \int_{0}^{1} y dy = 0.5$$

So the $E(Y|X)$ is given by:

$$E(Y|X) = \begin{cases} 1.5, & 0 \leq x \leq 0.5 \\
1, & 0.5 \leq x \leq 1 \\
0.5, & 1 \leq x \leq 1.5\end{cases}$$

  1. Compute the conditional expectation function of X given Y:

The conditional PDF of X given Y is given by:$$f_{X|Y}(x|y) = \frac{f(x, y)}{f_Y(y)}$$

Let's compute $f_Y(y)$ first:

For $0 \leq y \leq 1$:

$$f_Y(y) = \int_{0.5}^{1.5} 0.5 dx = 0.5$$

For $1 \leq y \leq 2$:

$$f_Y(y) = \int_{0}^{1} 0.5 dx = 0.5$$

So the $f_Y(y)$ is given by:

$$f_Y(y) = \begin{cases} 0.5, & 0 \leq y \leq 1 \\
0.5, & 1 \leq y \leq 2\end{cases}$$

And the $f_{X|Y}(x|y)$ is given by:

$$f_{X|Y}(x|y) = \begin{cases} 1, & 0 \leq y \leq 1\\
1, & 1 \leq y \leq 2\end{cases}$$

Now, let's compute the conditional expectation of X given Y:

For $0 \leq y \leq 1$:

$$E(X|Y) = \int_{0.5}^{1.5} x dx = 1$$

For $1 \leq y \leq 2$ is:

$$E(X|Y) = \int_{0}^{1} x dx = 0.5$$

So the $E(X|Y)$ is given by:

$$E(X|Y) = \begin{cases} 1, & 0 \leq y \leq 1 \\
0.5, & 1 \leq y \leq 2\end{cases}$$