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Let $ X \sim (0,1) $ and $ Y \sim (-1,2)$ be independent. Compute the distribution function of $Z=X+Y$ - how to break into cases?

I first found the density functions:

$$ f_x(t) =\begin{cases} 1 && t\in[0,1] \\ 0 && else \end{cases} $$

$$ f_y(t) =\begin{cases} \frac{1}{3} && t\in[-1,2] \\ 0 && else \end{cases} $$

Now:

$ F_z(t)=P(Z\leq t)=P(X+Y\leq t)=\int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$

But now I am stuck with breaking the result into cases.

How could it be done? I simply can't understand how to break that into cases when we have two differently distributed variables?

With one variable I would usually draw the function and then break to cases according to it's behavior, but how could it be done here?

Thanks

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  • $\begingroup$ when you wrote $z-y$, did you mean $t-y$? $\endgroup$ – Keen-ameteur Feb 11 at 15:17
  • $\begingroup$ @Keen-ameteur yes, my fault $\endgroup$ – superuser123 Feb 11 at 15:19
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    $\begingroup$ Use convolution theorem. $\endgroup$ – Dbchatto67 Feb 11 at 15:21
  • $\begingroup$ @Dbchatto67 Isn't there another way? I don't have in with convolutions. $\endgroup$ – superuser123 Feb 11 at 15:22
  • $\begingroup$ The range of $z$ is from -1 to 3. I think the distribution of $z$ can be written as a piece-wise continuous function in three parts, with the 'joins' at $z=0$ and $z=2$. $\endgroup$ – Paul Aljabar Feb 11 at 15:27
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By convolution theorem we get

$$f_Z(z) = \int_{0}^{1} f_X(x) f_Y(z-x)\ \text{dx}.$$

In order to make non vanishing integrand we should have $-1 \leq z-x \leq 2$ i.e. $z-2 \leq x \leq z+1$. Observe that $-1 \leq z \leq 3 .$ Now there are three cases to consider

$(1)$ $z-2 < 0 \leq z+1 < 1.$

$(2)$ $z-2 < 0 < 1 \leq z+1.$

$(3)$ $0 \leq z-2 \leq 1 < z+1.$

If you analyze each of these cases you will find that the probability density function $f_Z$ of $Z = X + Y$ is defined as follows $:$

$$f_Z(z) = \begin{cases} \frac 1 3(z+1) & \text{for $-1 \leq z < 0$} \\ \frac 1 3 & \text{for $0 \leq z < 2$} \\ \frac 1 3 (3-z) & \text{for}\ 2 \leq z \leq 3 \\ 0 & \text{elsewhere} \end{cases}$$

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First notice that for the inside integral to contribute something you must have:

$0 \leq t-y\leq 1$

Which means that:

$\int_{-1}^{2}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy= \int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy$

But you should notice that:

$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy\neq \int_{-1}^{2}\int_{0}^{t-y}\frac{1}{3}dxdy$

But instead: $t\leq -1$ then $\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$, and if $t>2$ $\int_{t}^{t-1}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=0$. else:

$\int_{t-1}^{t}\int_{0}^{t-y}f_X(x) f_Y(y)dxdy=\begin{cases}0 & ,t<-1 \\ 1 & ,t>2 \\ \int_{(t-1)\vee -1}^{t}\int_{0}^{t-y} \frac{1}{3} dxdy & ,\text{else} \end{cases}$

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Assuming that you know that the sum of two independent random variables is their convolution, we have $$F_z(t)= \int_{-\infty}^{\infty}f_X(t-\tau)f_Y(\tau) \ d \tau = \frac{1}{3} \int_{-1}^{2}f_X(t-\tau) \ d \tau = \frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx$$ Now, we have cases according to the integral boundaries and the PDF of $X$, namely:

1) If $t+1<0$, i.e. $t<-1$, the integral evaluates to zero.

2)If $t-2>1$, i.e. $t>3$, the integral evaluates to zero.

3) If $t-2<0$ ($t<2$) given that $t+1<1$, i.e. $t<0$ So for $-1 < t < 0$ $$F_z(t)=\frac{1}{3}\int_{0}^{t+1}f_X(x) \ dx = \frac{1}{3}(t+1)$$

4) If $t-2 > 0$ ($t > 2$). So for $2 <t < 3$

$$F_z(t)=\frac{1}{3}\int_{t-2}^{1}f_X(x) \ dx = \frac{1}{3}(3-t)$$

5) Now $0 < t < 2$

$$F_z(t)=\frac{1}{3}\int_{t-2}^{t+1}f_X(x) \ dx =\frac{1}{3}\int_{0}^{1} 1 \ dx =\frac{1}{3}$$

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