Let $$ X $$ and $$ Y $$ be two jointly continuous random variables with pdf $$ f_{x, Y}(x, y)=\left\{\begin{array}{ll}6 x y & \text { if } 0 \leq x \leq 1, \quad 0 \leq y \leq \sqrt{x} \ 0 & \text { otherwise }\end{array}\right. $$ a. Find $$ f_{x}(x) $$ and $$ f_{y}(y) $$ b. Find the conditional pdf of $$ X $$ given $$ Y=y, f_{x \mid Y}(x \mid y) $$

Question

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### Problem b

**Find the conditional probability density function $$ f_{X \mid Y}(x \mid y) $$.**

To find the conditional pdf $$ f_{X \mid Y}(x \mid y) $$, we use the definition:

$$
f_{X \mid Y}(x \mid y) = \frac{f_{X,Y}(x, y)}{f_Y(y)}
$$

From part **a**, we have:

1. **Joint PDF:**
   $$
   f_{X,Y}(x, y) = 
   \begin{cases}
   6xy & \text{if } 0 \leq x \leq 1 \text{ and } 0 \leq y \leq \sqrt{x} \
   0 & \text{otherwise}
   \end{cases}
   $$

2. **Marginal PDF of $$ Y $$, $$ f_Y(y) $$:**

To find $$ f_Y(y) $$, we integrate the joint PDF over all possible values of $$ x $$:

$$
   f_Y(y) = \int_{x = y^2}^{1} 6xy \, dx
   $$
   
   **Explanation of Limits:**
   - Given $$ 0 \leq y \leq \sqrt{x} $$, for a fixed $$ y $$, $$ x $$ must satisfy $$ y \leq \sqrt{x} $$, which implies $$ x \geq y^2 $$.
   - Since $$ x \leq 1 $$, the upper limit is 1.

**Performing the Integration:**
   $$
   f_Y(y) = 6y \int_{x = y^2}^{1} x \, dx = 6y \left[ \frac{x^2}{2} \right]_{y^2}^{1} = 6y \left( \frac{1}{2} - \frac{y^4}{2} \right) = 3y (1 - y^4)
   $$
   
   So,
   $$
   f_Y(y) = 
   \begin{cases}
   3y(1 - y^4) & \text{if } 0 \leq y \leq 1 \
   0 & \text{otherwise}
   \end{cases}
   $$

3. **Conditional PDF $$ f_{X \mid Y}(x \mid y) $$:**

Now, substitute $$ f_{X,Y}(x, y) $$ and $$ f_Y(y) $$ into the conditional PDF formula:

$$
   f_{X \mid Y}(x \mid y) = \frac{6xy}{3y(1 - y^4)} = \frac{2x}{1 - y^4}
   $$

**Valid Range:**
   - From the joint PDF, $$ y^2 \leq x \leq 1 $$ (since $$ 0 \leq y \leq \sqrt{x} $$ and $$ x \leq 1 $$).
   - Also, $$ 0 \leq y \leq 1 $$ because $$ y \leq \sqrt{1} = 1 $$.

Therefore,

$$
   f_{X \mid Y}(x \mid y) = 
   \begin{cases}
   \dfrac{2x}{1 - y^4} & \text{if } y^2 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \
   0 & \text{otherwise}
   \end{cases}
   $$

### Summary

The conditional probability density function of $$ X $$ given $$ Y = y $$ is:

$$
f_{X \mid Y}(x \mid y) = 
\begin{cases}
\dfrac{2x}{1 - y^4} & \text{for } y^2 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \
0 & \text{otherwise}
\end{cases}
$$
The conditional probability density function of $$ X $$ given $$ Y = y $$ is: $$ f_{X \mid Y}(x \mid y) = \begin{cases} \dfrac{2x}{1 - y^4} & \text{for } y^2 \leq x \leq 1 \text{ and } 0 \leq y \leq 1 \ 0 & \text{otherwise} \end{cases} $$

Let and be two jointly continuous random variables with pdf

a. Find and
b. Find the conditional pdf of given

Upstudy AI Solution

Answer

Solution

Problem b

Summary

Bonus Knowledge

a. Finding and

b. Finding the conditional pdf of given

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Let and be two jointly continuous random variables with pdf a. Find and b. Find the conditional pdf of given