Finding convolution of exponential distribution

Question

So this is a probability question, and I am asked to find $P(0.6 < Y <= 2.2)$

where $Y = X_1 + X_2$

X1~U(0,1) and X2~exp(2). Our professor worked it out, but I do not understand his explanation. So I understand that to find the probability of a sum of independent random variables, I need to use the convolution:

$$f_y (t) = \int_{-\infty}^\infty f_{x_1}(u)f_{x_2}(t-u) \, du $$

I know the density function ($f_{x_1})$ for a uniform distribution is 0 if u is less than 0 or greater than 1, so this whole integral is 0 except when u is between 0 and 1.

$$f_y (t) = \int_{0}^1 f_{x_1}(u)f_{x_2}(t-u) \, du $$, in which case $f_{x_1}$ is just 1, so we just have

$$f_y (t) = \int_{0}^1 f_{x_2}(t-u) \, du $$

We substitute y = t-u and du = -dy, and since we substituted, I know we change the limits of integration and now we have:

$$f_y (t) = \int_{t-1}^t f_{x_2}(y) \, dy $$

so for $$f_{x_2}(y) = \begin{cases} 0 & \text{if $x < 0$} \\ \lambda e^{-\lambda y} & \text{if $x>=0$} \end{cases}$$. I understand until this point, but at this point my professor

"divides it into cases":

case: (0 <= t <= 1)..

$$\int_{0}^t \lambda e^{-\lambda y} \, dy$$, also changing the limits of integration, and then in the case of (t > 1),

$$\int_{t-1}^t \lambda e^{-\lambda y} \, dy $$,

and then solves the integrals from there. I have NO CLUE why he divided into those "cases", and how he determined to what the limits of integration should be for each case, so if anyone could help me (I know this was a long problem), please !!

Answer 1

Let us formulate the cases in a different way:

1. If $0 \leq t \leq 1$ we have $t - 1 \leq 0 \leq t$ and then $$ \int_{t - 1}^t f_{x_2}(y) dy = \int_{t - 1}^0 f_{x_2}(y) dy + \int_{0}^t f_{x_2}(y) dy$$ and the first sum vanishes as $f_{x_2}(y) = 0$ for $y < 0$.

2. If $t > 1$ then $t - 1 > 0$ and so $$ \int_{t - 1}^t f_{x_2}(y) dy = \int_{t-1}^t \lambda e^{-\lambda y}$$

You can think about it this way: we are integrating over the interval $[t-1,t]$ and our function is zero outside of $[0,\infty)$, so we integrate over the interval $[t-1,t] \cap [0,\infty) = [\max(t-1,0),t]$ (and here we may replace $f_{x_2}(y)$ by $\lambda e^{-\lambda} y$).

Answer 2

Assume that $X_1$ and $X_2$ are independent and consider, for every $x$ in $(0,1)$, $$u(x)=P(0.6\leqslant x+X_2\leqslant2.2)=P((0.6-x)^+\leqslant X_2\leqslant2.2-x).$$ The density of $X_2$ is $2\mathrm e^{-2y}$ on $y\geqslant0$ hence $$u(x)=\int_{(0.6-x)^+}^{2.2-x}2\mathrm e^{-2y}\,\mathrm dy=\left[-\mathrm e^{-2y}\right]_{(0.6-x)^+}^{2.2-x}, $$ that is, $$u(x)= \begin{cases}\mathrm e^{-2(0.6-x)}-\mathrm e^{-2(2.2-x)} & \text{if} & 0\lt x\lt0.6\\ 1-\mathrm e^{-2(2.2-x)} & \text{if} & 0.6\lt x\lt1\end{cases}$$ Furthermore, $X_1$ is uniform on $(0,1)$ hence, by independence, $$P(0.6\leqslant X_1+X_2\leqslant2.2)=\int_0^1u(x)\mathrm dx,$$ which leads quickly to the result.

To prove the key formula above in a more general setting, consider some independent $X$ and $Y$ and note that, for every Borel set $B$, Tonelli theorem yields $$P((X,Y)\in B)=\iint\mathbf 1_{(x,y)\in B}\mathrm dP_X(x)\mathrm dP_Y(y)=\int u(x)\mathrm dP_X(x),\qquad u(x)=\int\mathbf 1_{(x,y)\in B}\mathrm dP_Y(y).$$