\[ f(x, y) = c (x^2 -\ y^2) \hspace{0.5in} 0 \leq x \lt \infty, \hspace{0.2in} -x \leq y \leq x \]

This question aims to find the **conditional distribution** of the given **function** with a given **condition** X=x.

The question is based **on the joint density function** and **conditional distribution** concepts. The conditional distribution is the probability of an item randomly selected from a population with some characteristics we want.

## Expert Answer

We are given a **function** f(x, y), which is **joint density function** with x and y limits. To find the **conditional distribution** of the joint **density function** with the given condition X=x, we first need to find the **marginal density** of X. The **marginal density** of X is given as:

\[ f_X(x) = \int_{-x}^{x} f(x, y) \, dy \]

\[ \int_{-x}^{x} f(x, y) \, dy = \int_{-x}^{x} c(x^2 -\ y^2) e^{-x} \, dy \]

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \int_{-x}^{x} (x^2 -\ y^2) \, dy \]

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \bigg{[} yx^2 -\ \dfrac{y^3}{3} \bigg{]}_{y=-x}^{y=x} \]

Substituting the value of $y$, we get:

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \bigg{[} \Big\{ \big{(} (x)x^2 -\ \dfrac{x^3}{3} \big{)} -\ \big{(} (-x)x^2 -\ \dfrac{-x^3}{3} \big{)} \Big\} \bigg{]} \]

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \bigg{[} \Big\{ \dfrac{3x^3 -\ x^3}{3} -\ \dfrac{-3x^3 + x^3}{3} \Big\} \bigg{]} \]

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \big{[} \dfrac{2x^3}{3} -\ \dfrac{-2x^3}{3} \big{]} \]

\[ \int_{-x}^{x} f(x, y) \, dy = c e^{-x} \big[ \dfrac{4x^3}{3} \big] \]

\[ f_X(x) = \dfrac{4c e^{-x} x^3}{3} \]

We can now find the **conditional distribution** of $Y$ with the given condition $X=x$ by using the following formula:

\[ f_{ Y|X }( y|x ) = \dfrac{f(x, y)} {f_X (x)} \]

\[ f_{ Y|X } ( y|x) = \dfrac{c (x^2 -\ y^2) e^{-x}} { \dfrac{ 4c e^{-x} x^3}{3}} \]

\[ f_{ Y|X } ( y|x) = \dfrac{ 3c e^{-x} (x^2 -\ y^2)} {4c e^{-x} x^3}\]

The **constants** $c$ and $e^{-x}$ will cancel each other and we get:

\[ f_{ Y|X } ( y|x) = \dfrac{ 3 (x^2 -\ y^2)} {4x^3}\hspace{0.5in} for\ x \gt 0 \hspace{0.2in} and\ -x \leq y \leq x \]

## Numerical Result

The **conditional distribution** of **function** $Y$ with given condition $X=x$ is calculated to be:

\[ f_{ Y|X } ( y|x) = \dfrac{ 3 (x^2 -\ y^2)} {4x^3} \]

## Example

Find the **marginal density function** of $X$ for the given **joint probability density function.**

\[ f(x) = c e^{-x} \dfrac{x^2}{2} \hspace{0.5in} -y \leq x \leq y \]

The **joint probability density function** is given, which is equal to $1$ as the **total probability** of any **density function.**

To solve for the **marginal density function,** we **integrate** the **function** over the given **limits** of $x$ as:

\[ f(x) = \int_{-y}^{y} \dfrac{c e^{-x} x^2} {2} \, dx \]

\[ f(x) = \dfrac{c e^{-x}} {2} \Big[ x^2 +2x +2 \Big]_{-y}^{y} \]

By substituting the values of limits in to the equation, we get:

\[ f(x) = \dfrac{c e^{-x}} {2} (2 y^2 + 2) \]

\[ f(x) = c e^{-x} (y + 1) \]