Question

The random variables X and Y are described by a uniform joint PDF of the form fx,y (x, y) = 3 on the set {(x, y) 0 < x ≤ 1, 0 ≤ y ≤ 1, y ≤ x²}.

Then, fx (0.5) =

Answer 1

To find the marginal probability density function (PDF) \( f_X(x) \) of the random variable \( X \), we need to integrate the joint PDF \( f_{X,Y}(x,y) \) over the appropriate range of \( Y \).

Given that the joint PDF \( f_{X,Y}(x,y) = 3 \) on the region defined by \( 0 < x \leq 1 \), \( 0 \leq y \leq 1 \), and \( y \leq x^2 \), we first need to determine the limits for \( y \) when we are conditioning on \( X = x \).

For a fixed \( x \), \( y \) can vary from \( 0 \) to \( x^2 \) (according to the constraints of the region). Therefore, the marginal PDF of \( X \) is calculated as follows:

\[
f_X(x) = \int_{y=0}^{y=x^2} f_{X,Y}(x,y) \, dy
\]

Substituting \( f_{X,Y}(x,y) \):

\[
f_X(x) = \int_{0}^{x^2} 3 \, dy
\]

Calculating this integral:

\[
f_X(x) = 3 \int_{0}^{x^2} dy = 3 [y]_{0}^{x^2} = 3(x^2 - 0) = 3x^2
\]

Thus, the marginal PDF of \( X \) is:

\[
f_X(x) = 3x^2 \quad \text{for } 0 < x \leq 1
\]

Now, we can find \( f_X(0.5) \):

\[
f_X(0.5) = 3(0.5^2) = 3(0.25) = 0.75
\]

Therefore, the value of \( f_X(0.5) \) is

\[
\boxed{0.75}
\]