Given the joint PDF:
$$f_{X,Y}(x,y) = \begin{cases} x+y-1\over5 &, -1 \le x \le 1, 1\le y \le 2 \\ 0 &, \text{otherwise}\end{cases}$$
I need to find:
a) the conditional PDF $f_{Y|X=0}(y)$
b) the conditional probability $P(Y \le 1.5 | X = 0)$
My solution:
I've solved a) by first computing the marginal PDF of Y, then the probability $P[X = 0] = 1 – P[X \neq 0] = {4\over5}$. And finally $f_{Y|X=0}(y) = {{y-1}\over2}$. As for b) I'm not really sure how to solve it :/. Any thoughts?
Best Answer
Please note that there is a mistake in your working of part $(a)$.
Marginal pdf of $X$, $f_X(x) = \displaystyle \int_1^2 \frac{x+y-1}{5} \ dy = \frac{2x+1}{10}$
$f_{Y|X}(y | x = 0) = \dfrac{f(0,y)}{f_X(0)} = \dfrac{(y-1)/5}{1/10} = 2 (y-1)$
So for (b),
$\displaystyle P(Y \le 1.5 | X = 0) = \int_1^{1.5} 2(y-1) \ dy$
Just as a side note - to validate that the conditional density you came up with is correct or not, you can evaluate
$\displaystyle \int_1^{2} f_{Y|X}(y | x = 0) = \int_1^{2} 2(y-1) \ dy$
and it should evaluate to $1$.