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Let $X,Y \sim Exp(1)$ be independent random variables. Let $W = X+Y$ and $Z = X / (X + Y)$ .

I am tasked with calculating $P(\{ Z < z \} \cap \{ W < w \}$).

I have the solution that says

enter image description here

So the solution says $$P \Big( \frac{X}{X+Y}

Where I am using 1 as the indicator function, and then calculates the integral.

The fact is I have never proven this equality for random vectors and I don't understand why the resulting expectation of indicator random variable should have the form above where we simply multiply together the densities of $X$ and $Y$.

So could someone please give me a link to a proof of the probability-expectation inequality for a random vector and try to explain to me why the expectation would be calculated that way?

I think a thing that confuses me is that the indicator function is on a set that is determined by $f(X,Y)$ instead of $X$ and $Y$ separately.

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    In the solution the probability is calculated [by definition](https://www.statlect.com/glossary/absolutely-continuous-random-vector). And from that it is derived that the expectation of the indicator is the probability of the event, not the other way around. As for the multiplication of densities, that is because X and Y are independent, so the joint density is the product of densities. – 2017-02-08
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    @Blaza thank you I now realize why the density is the product, but why would we not need first to calculate the joint density of $X / (X+Y)$ and $X+Y$, why can we use the joint density of $X,Y$ ? – 2017-02-09
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    You can do that because you're still calculating the probability that the vector $(X,Y)$ is in some set. That set is written a bit awkwardly. you should be able to rewrite it a bit to make it clearer what is the set with regard to $x$ and $y$. It's kinda like in the univariate case you could have to calculate $P(X<4-X, 2X>1+X)$ so you would integrate the pdf over the set $\{x\ :\ x<4-x, 2x>1+x\}$, but that boils down to $\{x\ :\ 1$4-X$ variable, for example. I hope this clears things a bit. – 2017-02-09
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    First of all you explain very well so thanks again. I still have a doubt though, In your univariate example I could always explicit the random variable $X$ and then apply the integration over the pdf. In my example you can't explicit either $X$ or $Y$. In your example I can't write $4-x$ in the upper limit of integration and get the same result than having $2 $ in the upper limit of integration. Did I make my doubt clear? – 2017-02-09
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    I understand, but after some manipulation, the set in your example becomes $\{x,y > 0 \ :\ \frac{1-z}zx$\int_0^{\infty}(\int_{\frac{1-z}zx}^{w-x}e^{-x}e^{-y}dy)dx$, which you can actually calculate. It doesn't matter that you can't limit both $X$ and $Y$ only by constants. – 2017-02-09

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