Question:
Given independent random variables $X$ and $Y$, and the corresponding density functions are $f_X(x)= \frac{1}{2\sqrt{x}} \text{, } x \in [0,1]$ and $f_Y(y)=\frac{1}{2\sqrt{y}} \text{, } y \in [0,1]$. If $S=X+Y$, calculate $P(S \leq 1)$.
What I've tried:
Using convolution probability distribution rules, we have:
$$f_S(s) = \int \frac{1}{2\sqrt{t}} \frac{1}{2\sqrt{s-t}}dt=?$$
I am not sure about the supports of this integral? By looking at density functions of $X$ and $Y$, we know that $x,y \in [0,1]$, does that imply the supports of $f_S$ are $0,1$, that is, $\int_0^1$?
Also, how to solve such integral? I know how to solve $\int \frac{1}{2\sqrt{t}}dt$, but have never deal with a multiplication involved in integral.
I would compute this via the following.
Let $X$ and $Y$ be independent random variables with pdf's $f_X(x)= \frac{1}{2\sqrt{x}}$ and $f_Y(y)= \frac{1}{2\sqrt{y}}$ respectively. Then, the joint distribution would be: $$f(x,y) = \frac{1}{\left(2 \sqrt{x}\right) \left(2 \sqrt{y}\right)}.$$ We wish to find the density of $S=X+Y$. One way to do this is to find $P(S\leq s)=P\left(X+Y\leq s\right)$ (i.e., the cdf), which turns out to be
$$\begin{array}{cc} F(s) = \left \{ \begin{array}{cc} 1 & s\geq 2 \\ \frac{\pi s}{4} & s\leq 1 \\ \sqrt{s-1}+\frac{1}{2} z \left(\csc ^{-1}\left(\sqrt{s}\right)-\tan ^{-1}\left(\sqrt{s-1}\right)\right) & 1 < s < 2\\ \end{array} \right. \\ \end{array}.$$
Differentiating, we get the density
$$\begin{array}{cc} f(s) = \left \{ \begin{array}{cc} \frac{\pi }{4} & s<1 \\ \frac{1}{2} \left(\frac{1}{2 \sqrt{z-1}}-\frac{1}{2 \sqrt{1-\frac{1}{s}} \sqrt{s}}-\tan ^{-1}\left(\sqrt{s-1}\right)+\csc ^{-1}\left(\sqrt{s}\right)\right) & 1<s<2 \\ \end{array} \\ \right. \end{array}$$
We can verify the density is correct by integrating over the domain of support and make sure it sums to 1, which it does as shown below.
$$\int_0^1 \frac{\pi }{4} \, dz+\int_1^2 \frac{1}{2} \left(\frac{1}{2 \sqrt{z-1}}-\frac{1}{2 \sqrt{1-\frac{1}{z}} \sqrt{z}}-\tan ^{-1}\left(\sqrt{z-1}\right)+\csc ^{-1}\left(\sqrt{z}\right)\right) \, dz = 1$$
To compute $P(S \leq 1)$, we can use the cdf or the pdf - whatever. It turns out that $$P(S \leq 1) = \frac{\pi}{4}$$ as you can see by examining the cdf or integrating the pdf.