1
$\begingroup$

Two independent random variable X,Y which is U(0,1)

what is the pdf of Z=X+Y?

$F_Z(z)$ = \begin{cases} 0, & \ z\le0 \\ \int_0^z \int_0^{z-y} 1 \,dxdy=z^2/2, & \ 0\lt z\le1 \\ 1-\int_\bbox[yellow]{z-1}^1 \int_{z-y}^1 1 \,dxdy=1-(2-z)^2/2,& \ 1\lt z\le2 \\ 1 & \ z\ge2 \end{cases}

I understand I have to differential $F_z(z)$

The thing I can't understand is the highlight part.

Why does it starts with z-1?

$\endgroup$
  • 1
    $\begingroup$ By union, did you mean uniform? The usual notation for this distribution is $U(0,\,1)$. $\endgroup$ – J.G. Jan 5 at 14:25
  • $\begingroup$ @J.G. Thank you I edited it. $\endgroup$ – yonghankwon0 Jan 6 at 5:01
1
$\begingroup$

Convolution of probability densities

Let $X$ and $Y$ be any two independent (real-valued) random variables with densities $f_X$ and $f_Y$. Define $Z \equiv X + Y$. Note that $$ F_Z(z) = \mathbb{P}(Z \leq z) = \mathbb{P}(X + Y \leq z) = \int_{-\infty}^{\infty} f_X(x) \mathbb{P}(x + Y \leq z) dx = \int_{-\infty}^{\infty} f_X(x) F_Y(z - x) dx. $$ Differentiating, $$ f_Z(z) = \int_{-\infty}^{\infty} f_X(x) f_Y(z - x) dx. $$


Convolution of i.i.d. uniform distributions

If $X$ and $Y$ are i.i.d. $U(0,1)$, then $f_x = f_Y = \boldsymbol{1}_{(0,1)}$ where $\boldsymbol{1}$ is the indicator function. Plugging this into the integral of the previous section, $$ f_Z(z) = \int_0^1 \boldsymbol{1}_{(0,1)}(z - x) dx. $$ Proceeding by cases, $$ f_Z(z) = \begin{cases} \int_0^z dx = z & \text{if } 0 \leq z \leq 1 \\ \int_\bbox[yellow]{z-1}^1 dx = 2 - z & \text{if } 1 \leq z \leq 2. \\ \end{cases} $$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.