https://file.aiquickdraw.com/mathbot/file/1734551833454yen0t6j3.jpg

https://file.aiquickdraw.com/mathbot/file/1734551833455u5nowsn1.jpg

### Problem Recap
You are given the **joint density** \( f_{X,Y}(x,y) \) of two jointly continuous random variables \( X \) and \( Y \):

\[
f_{X,Y}(x,y) =
\begin{cases}
e^{-(x+y)}, & x > 0, \, y > 0 \\
0, & \text{otherwise}.
\end{cases}
\]

You need to:
1. Compute the **marginal density of \( Y \)**, \( f_Y(y) \).
2. Compute the **conditional density of \( X \) given \( Y \)**, \( f_{X|Y}(x|y) \).

---

### Part (i): Marginal Density of \( Y \), \( f_Y(y) \)

The marginal density of \( Y \) is obtained by integrating the joint density \( f_{X,Y}(x,y) \) over all values of \( x \):

\[
f_Y(y) = \int_{0}^{\infty} f_{X,Y}(x,y) \, dx.
\]

Substitute \( f_{X,Y}(x,y) = e^{-(x+y)} \) for \( x > 0 \) and \( y > 0 \):

\[
f_Y(y) = \int_{0}^{\infty} e^{-(x+y)} \, dx.
\]

The exponential term can be split:

\[
f_Y(y) = \int_{0}^{\infty} e^{-x} e^{-y} \, dx.
\]

Factor out \( e^{-y} \) (since it does not depend on \( x \)):

\[
f_Y(y) = e^{-y} \int_{0}^{\infty} e^{-x} \, dx.
\]

The integral of \( e^{-x} \) over \( x \in [0, \infty) \) is 1:

\[
\int_{0}^{\infty} e^{-x} \, dx = 1.
\]

Thus, the marginal density of \( Y \) is:

\[
f_Y(y) = e^{-y}, \quad y > 0.
\]

---

### Part (ii): Conditional Density of \( X \) Given \( Y \), \( f_{X|Y}(x|y) \)

The conditional density \( f_{X|Y}(x|y) \) is defined as:

\[
f_{X|Y}(x|y) = \frac{f_{X,Y}(x,y)}{f_Y(y)}.
\]

Substitute the known expressions for \( f_{X,Y}(x,y) \) and \( f_Y(y) \):

- \( f_{X,Y}(x,y) = e^{-(x+y)} \) for \( x > 0 \), \( y > 0 \),
- \( f_Y(y) = e^{-y} \) for \( y > 0 \).

Thus:

\[
f_{X|Y}(x|y) = \frac{e^{-(x+y)}}{e^{-y}}.
\]

Simplify the expression:

\[
f_{X|Y}(x|y) = e^{-x}, \quad x > 0, \, y > 0.
\]

The conditional density \( f_{X|Y}(x|y) \) is therefore:

\[
f_{X|Y}(x|y) =
\begin{cases}
e^{-x}, & x > 0, \, y > 0, \\
0, & \text{otherwise}.
\end{cases}
\]

---

### Final Answers:
1. **Marginal density of \( Y \):**
   \[
   f_Y(y) = e^{-y}, \quad y > 0.
   \]
2. **Conditional density of \( X \) given \( Y \):**
   \[
   f_{X|Y}(x|y) = e^{-x}, \quad x > 0, \, y > 0.
   \]

Let me know if you would like detailed explanations for each step!

---

### Related Questions:
1. How do you find the marginal density of \( X \)?
2. What is the expected value of \( X \) given \( Y = y \)?
3. How do you compute the joint cumulative distribution function \( F_{X,Y}(x,y) \)?
4. What is the covariance between \( X \) and \( Y \)?
5. How do you check if \( X \) and \( Y \) are independent random variables?

### Tip:
To check if two random variables \( X \) and \( Y \) are independent, verify if their joint density can be factored into the product of their marginal densities.

Let X and Y be jointly continuous random variables with the joint density function:

f_{X,Y}(x,y) = { e^{-(x+y)}, x > 0, y > 0 
                 0, otherwise }

(i) Compute the marginal density of Y, denoted f_Y(y). (5 points)
(ii) Compute the conditional density of X given Y, f_{X|Y}(x|y). (5 points)

Learn how to compute the marginal density and conditional density for two jointly continuous random variables with a joint density function f_{X,Y}(x,y). Step-by-step solution included for calculating f_Y(y) and f_{X|Y}(x|y).

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation