Finding a marginal probability density (stuck at the integration)

[MathNugget]

Hello.
This is $f(x, y) = y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)$
Trying to find the marginal probability density of Y.
so I am supposed to solve:
$$f_Y(x, y)= \int_{-\infty}^{\infty}y x^{y-1}e^{-y}\mathbb{1}_{(0, \infty)}(y)\mathbb{1}_{(0, 1)}(x)dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{-\infty}^{\infty}x^{y-1} \mathbb{1}_{(0, 1)}(x) dx = ye^{-y}\mathbb{1}_{(0, \infty)}(y) \int_{0}^{1}x^{y-1} dx = ...$$now I suppose $\mathbb{1}_{(0, \infty)}(y)$ tells me y > 0, so
$$\int x^{y-1} dx = \frac{x^y}{y}$$$$\int_{0}^{1}x^{y-1} dx = \frac{1}{y}?$$
Is it $$f_Y(x, y)=ye^{-y}\mathbb{1}_{(0, \infty)}(y) \frac{1}{y}= e^{-y}\mathbb{1}_{(0, \infty)}(y)?$$

 

[blamocur]

Your integration result looks good to me.

 

[MathNugget]

1)
Ok, thanks.
I realize now, I could get rid of x in the last part and write $f_Y (y)$.

I hope it's not against the forum rules, but I have some further questions (I dont want to spam the forum, but they're only half-related as they're part of the same exercise):
(my choice of words to describe these statistics notions might be wrong, I am trying to double check with chat gpt):
If I want to simulate some random variables from this $f_Y$, I know 2 methods: one is to calculate F from f, and then its inverse, and then I have a formula, second is to find another function, g(y) , such that $f_Y(y)$<cg(y)...

I guess I can use the first method on this, but would the usual formula work here?
$$F_Y(y) = \int_{-\infty}^{x}f_Y(y)dy?$$I see $f_Y$ is very easy to integrate, so I am tempted to use the first method (if it can be applied)

 

[MathNugget]

I figured I can leave for questions here, it might be more effective:

2) How do I find the density of X with the condition that Y=y?
Is it: $f_{X | Y} (X=x | Y=y) = \frac{f_{X,Y}(x, y)}{f_Y(y)}$, and then I just do some simple calculations?

Furthermore, when calculating $\mathbb{P}(X \leq x | Y =y)$, do I just $\int_{-\infty}^{x}f_{X | Y} (X=x | Y=y) dx$