0

Consider random variable $X_i$s are independent and identically distributed. We assume that each $X_i$ is uniformly distributed in $[0, 1].$

(a) Find the cdf and pdf for $Z = X1 + X2$.

(b) Draw the pdf of $Z$.

So for (a), I understand that we should be first be finding the cdf of $Z$ by:

$F_Z(z)=P(X_1+X_1\le Z)=\int_{X_2=...}^{X_2=...}\int_{X_1=...}^{X_1=...} f(x_1,x_2)dx_1 dx_2$

and that $f(x_1,x_2)=1\times 1$

However, I'm really confused in how to set up the bounds:

$0 \le x_1 \le 1, 0 \le x_2 \le 1, x_1+x_2 \le z$

I tried doing from $\int_{X_2=0}^{X_2=1}\int_{X_1=0}^{X_1=z-x_2}$, but I now for sure that something is wrong since I got my pdf to have an area of 2. Could someone help?

Sami
  • 35
  • the formula and the limits are correct for the CDF. – Kphysics Nov 23 '19 at 21:24
  • I doubt it, because if you use those limits, you end up with an area under the pdf which equals to 2, which is impossible since it has be equal to 1. – Sami Nov 23 '19 at 21:26
  • Probably a lot easier to do via MGF if you are interested in learning about those.

    The limit on $x_1$ is not correct. $z-x_2$ can be greater than 1,

     – fGDu94 Nov 23 '19 at 21:29
  • $\int_0^1 \int_0^1 dX_1 dX_2 = 1 \neq 2$ – Kphysics Nov 23 '19 at 21:32
  • yeah but you want the CDF of Z as well, and this should be at most 1 – fGDu94 Nov 23 '19 at 21:33
  • Like this $F_Z(z) = \int_0^1 \int_0^{\max(min(z-x_2),1),0)}1 dx_1 dx_2$. It is somewhat clunky but does the job.

    I checked and it integrates to 1 over $z \in [0,2]$

     – fGDu94 Nov 23 '19 at 21:46

2 Answers2

2

We compute the density of $Z$ by convolution. Let $f_{X_1}(t)=\mathsf 1_{(0,1)}(t)$ and $f_{X_2}(t)=\mathsf 1_{(0,1)}(t)$ be the densities of $X_1$ and $X_2$, respectively. Then the density of $Z=X_1+X_2$ is given by \begin{align} f_Z(t) &= (f_{X_1}\star f_{X_2})(t)\\ &= \int_{\mathbb R}f_{X_1}(\tau)f_{X_2}(t-\tau)\ \mathsf d\tau.\\ &= \int_{\mathbb R}\mathsf 1_{(0,1)}(\tau)\mathsf 1_{(0,1)}(t-\tau)\ \mathsf d\tau. \end{align} For the integrand to be positive, we must have $0<\tau<1$ and $0<t-\tau<1$. First note that because $X_1$ and $X_2$ take values in $(0,1)$, $Z$ takes values in $(0,2)$. So for $0<t<1$ we have $0<\tau<t$, so that $$ f_Z(t) = \int_0^t \ \mathsf d\tau = t. $$ For $1<t<2$ we have $t-1<\tau<1$, so that $$ f_Z(t) = \int_{t-1}^1 \ \mathsf d\tau = 2 - t. $$ It follows that $$ f_Z(t) = t\cdot\mathsf 1_{(0,1)}(t) + (2-t)\cdot\mathsf 1_{(1,2)}(t). $$ We compute the distribution function of $Z$ by integrating the density. For $0\leqslant t<1$ we have $$ F_Z(t) = \int_0^t s\ \mathsf ds = \frac12 t^2. $$ For $1\leqslant t<2$ we have \begin{align} F_Z(t) &= \int_0^1 s\ \mathsf ds + \int_1^t (2-s)\ \mathsf ds\\ &= \frac12 + \frac12(t-1)(3-t)\\ &= \frac12 (1 +(t-1)(3-t)). \end{align} It follows that $$ F_Z(t) = \begin{cases} 0,&t<0\\ \frac12t^2,&0\leqslant t<1\\ \frac12 (1 +(t-1)(3-t)),&1\leqslant t<2\\ 1,&t\geqslant 2. \end{cases} $$

Math1000
  • 36,983
0

The pdf of Z can be determined by convoying the pdf of X1 with the pdf of X2. By the convolution theorem, the result equals the inverse Fourier Transform of the multiplication of their characteristic functions.

The two PDFs are shifted rect functions. Therefore, their characteristic functions is a sinc (with a linear phase). Therefore, the multiplication is sinc^2, with appropriate phase factor. Therefore the pdf of Z is a tri (shifted appropriately).

The tri function will be centered at 1.

NicNic8
  • 6,951