0
$\begingroup$

Let the joint density function of the random variables $X$ and $Y$ be

$$ f(x,y) = \left\{ \begin{array}{ll} 2xe^{x^2-y} & \quad 0 < x < 1, y > x^2 \\ 0 & \quad otherwise \end{array} \right. $$

(a) Find the marginal densities of $X$ and $Y$.

(b) Compute $P(Y < 3X^2)$.

(c) Find the conditional density function $f_{Y|X}(y|x)$.

(d) Find the conditional probability $P(Y \geq \frac{1}{4} | X = x)$ and verify the averaging identity $P(Y \geq \frac{1}{4}) = \int_{-\infty}^{\infty} P(Y \geq \frac{1}{4} | X = x) f_X(x)dx$.

First, I'd like to verify my solutions to parts (a) - (c) :

(a) $f_X(x) = \int_{x^2}^{\infty} 2xe^{x^2-y}dy = -2xe^{x^2-y} \Big|_{y = x^2}^{y = \infty} = 2xe^{x^2 - x^2} = 2x$ $\quad$ ($0 < x < 1$)

$f_Y(y) = \int_{0}^{\sqrt{y}} 2xe^{x^2-y}dx = e^{x^2-y} \Big|_{x=0}^{x = \sqrt{y}} = 1-e^{-y}$ $\quad$ ($0 < y < \infty$)

(b) $P(Y < 3X^2) = \int_{x^2}^{3x^2} (1-e^{-y})dy = (y + e^{-y}) \Big|_{y = x^2}^{y = 3x^2} = 2x^2 + e^{-3x^2} - e^{-x^2}$

(c) $f_{Y | X}(y | x) = \frac{f(x,y)}{f_X(x)} = \frac{2xe^{x^2-y}}{2x} = e^{x^2-y}$ $\quad$ ($x^2 < y < \infty$)

Is all of this correct ? The correct answers to parts (a)-(c) are critical to part (d), which is the part I'm really struggling with -- most likely because I've made a mistake in parts (a)-(c).

(d) First, we compute the left-hand side of the desired equality.

$P(Y \geq \frac{1}{4}) = \int_{\frac{1}{4}}^{\infty} (1-e^{-y})dy = (y + e^{-y}) \Big|_{\frac{1}{4}}^{\infty} $

But, this is where I know I've made a mistake, because this gives a probability of $\infty$. Have I made a crucial mistake somewhere ? Perhaps, I have the wrong marginal density function for $Y$ ?

Computing the right-hand side of the desired equality, I get

$\int_{-\infty}^{\infty} P(Y \geq \frac{1}{4} | X = x) f_X(x)dx = \int_{0}^{1} \int_{\frac{1}{4}}^{\infty} 2xe^{x^2-y}dydx = \int_{0}^1 2xe^{x^2 - \frac{1}{4}}dx = e^{x^2-\frac{1}{4}} \Big|_{0}^{1} = e^{\frac{3}{4}} - e^{\frac{-1}{4}} $

Is this computation at least correct ? If so, where have I gone wrong in computing the left-hand side ?

Thanks!

$\endgroup$
0

1 Answer 1

Reset to default
1
$\begingroup$

$f_Y(y)=\int_0^{\sqrt y} 2xe^{x^{2}-y}dx$ if $y <1$ and $f_Y(y)=\int_0^{1} 2xe^{x^{2}-y}dx$ if $y >1$.

$\endgroup$
2
  • $\begingroup$ Hello ! Thanks for the comment. I believe this still leads to the same issue in part (d) -- since $f_Y(y) = 1-e^{-y}$ for $y > 1$ and $f_Y(y) = e^{1-y} - e^{-y}$ when $y < 1$, computing the left-hand side of the desired equality, we now get $P(Y \geq \frac{1}{4}) = \int_{\frac{1}{4}}^1 (e^{1-y} - e^{-y})dy + \int_1^{\infty} (1-e^{-y}) dy$. Still, the second integral is an issue. $\endgroup$
    – michiganbiker898
    Feb 20, 2020 at 13:04
  • 1
    $\begingroup$ The first expression is for $y<1$ and the second for $y >1$. You are mixing things up. Please check your calculations again. $\endgroup$
    – Kavi Rama Murthy
    Feb 20, 2020 at 13:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .