Calculate $P(X > 1\mid Y = y)$ for both $x, y > 0$

Question

The joint density is:

$$f(x,y)=\frac{e^{-x/y}e^{-y}}{y}$$ Calculate $P(X > 1\mid Y = y)$

I'm going to consider the conditional density: I know that:

$f(x\mid Y=y)=\frac {f_{X,Y}(x,y)}{f_Y(y)}$ Then I could use the fact that:

$P(X>1\mid Y=y)=\frac{P(X>1,Y=y)}{P(Y=y)}$

where:

$$P(X>1,Y=y)= \int_{-\infty}^\infty P(X>1\mid Y=y)f_Y(y)dy = \int_{-\infty}^\infty \int_1^\infty f(x\mid Y=y)\,dx f_Y(y)\,dy=\int_{-\infty}^\infty \int_1^\infty f(x,y) \, dy \, dx$$

I know it's incorrect because I think the following is equal $P(X>1,Y=1)=P(X>1\mid Y=1)$ More questions,

is $f(x\mid Y=y)$ simply the probability density function for $x$, given a value of $y$?

Answer 1

Note that $$ f_Y(y) = \int_0^\infty f_{X,Y} (x,y) dx = \frac{e^{-y} }{y} \int_0^\infty e^{ -x/y } dx = e^{-y} $$ Thus you now know the conditional density, so $$ \mathbb{P}( X > 1 | Y = y ) = \int_1^\infty f_{X|Y} dx = \int_1^\infty \frac{ f_{X,Y} } { f_Y} dx = \frac{1}{y} \int_1^\infty e^{- x/y} dx = e^{-1/y} $$

Answer 2

As a hint you can notice that the joit pdf can be splitted into two functions, i.e., you have $f(x|y) = \frac{e^{-\frac{x}{y}}}{y}$ and $f(y) =e^{-y}$.

You can verify that by finding the marginal of y.

Then finding probability is straight forward. Hope this helps.