Question

The random variables X and Y are jointly continuous, with a joint PDF of the form

f X,Y(x,y)=c x y, if 0≤x≤ y ≤1, 0, otherwise,

where c is a normalizing constant.

For x€ [0,0.5], the conditional PDF f X given Y (x given 0.5) is of the form ax^b Find a and b Your answers should be numbers.
a

b

Answer 1

To find the conditional PDF \( f_{X|Y}(x | 0.5) \), we first need to determine the normalizing constant \( c \) for the joint PDF \( f_{X,Y}(x,y) = cx y \) in the specified region \( 0 \leq x \leq y \leq 1 \).

### Step 1: Find the normalizing constant \( c \)

The joint PDF must integrate to 1 over the specified region. We need to set up the double integral:
\[
\int_0^1 \int_0^y cxy \, dx \, dy
\]

First integrate with respect to \( x \):
\[
\int_0^y cxy \, dx = c y \int_0^y x \, dx = c y \left[ \frac{x^2}{2} \right]_0^y = c y \left( \frac{y^2}{2} \right) = \frac{c y^3}{2}
\]

Now, integrate with respect to \( y \):
\[
\int_0^1 \frac{c y^3}{2} \, dy = \frac{c}{2} \left[ \frac{y^4}{4} \right]_0^1 = \frac{c}{2} \cdot \frac{1}{4} = \frac{c}{8}
\]

Setting the integral equal to 1 for normalization:
\[
\frac{c}{8} = 1 \implies c = 8
\]

### Step 2: Determine the conditional PDF \( f_{X|Y}(x | 0.5) \)

We already have the joint PDF:
\[
f_{X,Y}(x,y) = 8xy, \quad \text{for } 0 \leq x \leq y \leq 1.
\]

Now we can find the marginal PDF \( f_Y(y) \):
\[
f_Y(y) = \int_0^y f_{X,Y}(x,y) \, dx = \int_0^y 8xy \, dx = 8y \int_0^y x \, dx = 8y \left[ \frac{x^2}{2} \right]_0^y = 8y \cdot \frac{y^2}{2} = 4y^3.
\]

The marginal PDF \( f_Y(y) \) is valid for \( 0 \leq y \leq 1 \).

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

For \( y = 0.5 \):
\[
f_Y(0.5) = 4(0.5)^3 = 4 \cdot \frac{1}{8} = \frac{1}{2}.
\]

Now compute \( f_{X,Y}(x, 0.5) \):
\[
f_{X,Y}(x, 0.5) = 8x(0.5) = 4x, \quad \text{for } 0 \leq x \leq 0.5.
\]

Thus, the conditional PDF \( f_{X|Y}(x | 0.5) \) becomes:
\[
f_{X|Y}(x|0.5) = \frac{4x}{\frac{1}{2}} = 8x, \quad \text{for } 0 \leq x \leq 0.5.
\]

### Step 3: Form \( ax^b \)

Comparing with the form \( ax^b \):
\[
f_{X|Y}(x|0.5) = 8x^1
\]
gives \( a = 8 \) and \( b = 1 \).

Thus, the answers are:
\[
\boxed{8} \quad \text{and} \quad \boxed{1}.
\]