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I'm struggling to understand the endpoints in this problem from Degroot (3.9.4 example).

Suppose also that $X_1$ and $X_2$ are independent random variables with common distribution having p.d.f. $f(x) = 2e^{−2x}$ for x > 0 and 0 otherwise

We're interested in $Y = X_1 + X_2$. In particular, we want to know the pdf of Y.

The solution is found by first finding the CDF:

$G(y) = Pr((X1, X2) ∈ Ay) = \int_0^{y} \int_0^{y−x_2} f(x)f(x)dx_1 dx_2$

However, this doesn't quite make sense. Rearranging such that:

$X_1 = Y - X_2$
$X_2 = Y - X_1$

I would think that to find the CDF we would integrate over:

$\int_0^{y - x_1} \int_0^{y−x_2} f(x)f(x)dx_1 dx_2$

Why do we only integrate over the range of $[0,y - x_1]$ in the outer integral?

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However, this doesn't quite make sense. Rearranging such that:

$X_1 = Y - X_2$
$X_2 = Y - X_1$

This is where you are mistaken, you don't have to "rearrange" twice. The proper way to go about it would be to write $$\mathbb P(Y\le y) = \mathbb P(X_1+X_2 \le y) = \mathbb P(X_1\le y-X_2) $$ And by the law of total probability we can rewrite this quantity as follows : $$\begin{align}\mathbb P(X_1\le y-X_2) &= \int_{x_2=0}^\infty\mathbb P(X_1\le y-X_2\ | \ X_2=x_2)f_{X_2}(x_2)dx_2\\ &= \int_{x_2=0}^y\mathbb P(X_1\le y-X_2\ | \ X_2=x_2)f_{X_2}(x_2)dx_2 \ \text{(the integrand is zero for $x_2\ge y$)}\end{align} $$ And similarly $$\begin{align}\mathbb P(X_1\le y-X_2\ | \ X_2=x_2) &= \mathbb P(X_1\le y-x_2)\\ &=\int_{x_1=0}^{y-x_2}f_{X_1}(x_1)dx_1\end{align} $$ Plug this in the previous inequality and the result follows.

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  • $\begingroup$ THANKS! What would be a better way to describe the "rearrangement" I did? I noticed you put in scare quotes. $\endgroup$ Feb 1, 2022 at 17:42
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    $\begingroup$ Ah no the quotation marks were not meant to say it's a wrong name for that operation, I was just quoting you ;) $\endgroup$ Feb 1, 2022 at 18:12

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