Question

Let Y,X have joint pdf
f xy( x,y)={ 4 ; 0 smaller than equal to x smaller than equal to 1 , 0 smaller than equal to y smaller than equal to 1
and 0 ; otherwise
Find fx ( x) , fy ( y ) ,fx/y ( x/y) ,f y/x ( y/x ) also find E ( X/ Y = y) and E (Y /X = x)

Answer 1

Answer on Question #65386 - Math - Statistics and Probability

**Question:** Let $Y, X$ have joint pdf

$$f_{XY}(x, y) = \begin{cases} 4, & 0 \leq x \leq 1, 0 \leq y \leq 1; \\ 0, & \text{otherwise} \end{cases}$$

Find $f_X(x), f_Y(y), f_{X/Y}(x/y), f_{Y/X}(y/x)$, also find $E(X / |Y = y)$ and $E(Y / |X = x)$.

**Solution:** First of all the given function is not joint pdf for any random variables $Y, X$. Indeed, $\int_{-\infty}^{+\infty} \left( \int_{-\infty}^{+\infty} f_{XY}(x, y) dy \right) dx = \int_{0}^{1} \left( \int_{0}^{1} 4 dy \right) dx = 4 \cdot 1 \cdot 1 = 4 \neq 1$.

Therefore, in order to obtain a joint pdf we have to change the constant 4 or the domain $0 \leq x \leq 1, 0 \leq y \leq 1$ in the description of $f_{XY}(x, y)$.

Let us consider such joint pdf

$$f_{XY}(x, y) = \begin{cases} a, & 0 \leq x \leq \frac{1}{\sqrt{a}}, 0 \leq y \leq \frac{1}{\sqrt{a}}; \\ & 0, \text{otherwise}, \end{cases}$$

where $a$ is an arbitrary positive real number.

The marginal density functions are

$$f_X(x) = \int_{-\infty}^{+\infty} f_{XY}(x, y) \, dy = \int_{0}^{1/\sqrt{a}} a \, dy = a y \bigg|_{0}^{1/\sqrt{a}} = \sqrt{a}, \text{ when } 0 \leq x \leq \frac{1}{\sqrt{a}},$$

and

$$f_Y(y) = \int_{-\infty}^{+\infty} f_{XY}(x, y) \, dx = \int_{0}^{1/\sqrt{a}} a \, dx = a x \bigg|_{0}^{1/\sqrt{a}} = \sqrt{a}, \text{ when } 0 \leq y \leq \frac{1}{\sqrt{a}}.$$

For $0 \leq y \leq \frac{1}{\sqrt{a}}$

$$f_{X/Y}(x/y) = \frac{f_{XY}(x,y)}{f_Y(y)} = \frac{a}{\sqrt{a}} = \sqrt{a}, \text{ when } 0 \leq x \leq \frac{1}{\sqrt{a}};$$

otherwise $f_{X/Y}(x/y) = 0$.

For $0 \leq x \leq \frac{1}{\sqrt{a}}$

$$f_{Y/X}(y/x) = \frac{f_{XY}(x,y)}{f_X(x)} = \frac{a}{\sqrt{a}} = \sqrt{a}, \text{ when } 0 \leq y \leq \frac{1}{\sqrt{a}};$$

otherwise $f_{Y/X}(y/x) = 0$.

Let us now calculate corresponding conditional expectations

$$E(X|Y = y) = \int_{-\infty}^{+\infty} x \, f_{X/Y}(x/y) dx = \int_{0}^{1/\sqrt{a}} \sqrt{a} \cdot x \, dx = \sqrt{a} \cdot \frac{x^2}{2} \bigg|_{0}^{1/\sqrt{a}} = \frac{1}{2\sqrt{a}}, \text{ when } 0 \leq y \leq \frac{1}{\sqrt{a}}.$$ $$E(Y|X = x) = \int_{-\infty}^{+\infty} y \, f_{Y/X}(y/x) dy = \int_{0}^{1/\sqrt{a}} \sqrt{a} \cdot y \, dy = \sqrt{a} \cdot \frac{y^2}{2} \bigg|_{0}^{1/\sqrt{a}} = \frac{1}{2\sqrt{a}}, \text{ when } 0 \leq x \leq \frac{1}{\sqrt{a}}.$$

**Answer:** For a joint pdf

$$f_{XY}(x, y) = \begin{cases} a, & 0 \leq x \leq \frac{1}{\sqrt{a}}, 0 \leq y \leq \frac{1}{\sqrt{a}}; \\ & 0, \text{otherwise}, \end{cases}$$ $$f_X(x) = \begin{cases} \sqrt{a}, & 0 \leq x \leq \frac{1}{\sqrt{a}}, \\ 0, & \text{otherwise}; \end{cases}, \quad f_Y(y) = \begin{cases} \sqrt{a}, & 0 \leq y \leq \frac{1}{\sqrt{a}}, \\ 0, & \text{otherwise}; \end{cases}$$

For $0 \leq y \leq \frac{1}{\sqrt{a}} f_{X/Y}(x/y) = \begin{cases} \sqrt{a}, & 0 \leq x \leq \frac{1}{\sqrt{a}}, \\ 0, & \text{otherwise}; \end{cases}$ and $E(X|Y = y) = \frac{1}{2\sqrt{a}}$.

For $0 \leq x \leq \frac{1}{\sqrt{a}} f_{Y/X}(y/x) = \begin{cases} \sqrt{a}, & 0 \leq y \leq \frac{1}{\sqrt{a}}, \\ 0, & \text{otherwise}; \end{cases}$ and $E(Y|X = x) = \frac{1}{2\sqrt{a}}$.

For marginal density function, conditional density function and conditional expectation see for example W. Feller, "An introduction to probability theory and its applications", 2, Wiley (1971), P.66-67 and 71-72. https://www.encyclopediaofmath.org/index.php/Feller_%22An_introduction_to_probability_theory_and_its_applications%22

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