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 !!

  • 1
    $\begingroup$ Two remarks. (1) One must assume that X1 and X2 are independent otherwise the solution is incorrect. (2) It is not necessary to "find the probability of a sum of independent random variables" to solve the question. $\endgroup$ – Did Dec 3 '14 at 8:04
  • $\begingroup$ @Did isn't that what's happening? $\endgroup$ – FrostyStraw Dec 3 '14 at 8:04
  • $\begingroup$ ?? What is happening? $\endgroup$ – Did Dec 3 '14 at 8:08
  • $\begingroup$ @Did Isn't that what the convolution is used for here? to like..find the probability of a sum of independent random variables? Like my professor went over a whole proof using the distribution function and at the end we ended up deriving the convolution, so what I got out of that is that that's whta you need to do to find the probability when you have a variable that is the sum of 2 indepdnent variables. Is that what his solution is doing --"finding the probability of a sum of independent random variables"? If not, what is he doing? And yeah, the problem assumes X1 and X2 are independent. $\endgroup$ – FrostyStraw Dec 3 '14 at 8:11
  • $\begingroup$ (1) If the problem assumes independence, you should write it in the question. (2) Yes convolution is needed to find the distribution of a sum of independent random variables. (3) No the distribution of X1+X2 is not needed to solve this exercise, see my answer. $\endgroup$ – Did Dec 3 '14 at 8:17

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 $$P((X,Y)\in B)=\iint\mathbf 1_{(x,y)\in B}\mathrm dP_X(x)\mathrm dP_Y(y)=\int u(x)\mathrm dP_X(x),\qquad u(x)=\int\mathbf 1_{(x,y)\in B}\mathrm dP_Y(y).$$

  • $\begingroup$ What is the + in $(0.6-x)^+$ ? I'm not sure how you got the piecewise function at the end, but it happens to be the same answer my professor got, so now I see two ways to do it and don't understand either lol @Did $\endgroup$ – FrostyStraw Dec 3 '14 at 8:28
  • $\begingroup$ $x^+$ is the positive part of $x$, that is, $x^+=x$ if $x\geqslant0$ and $x^+=0$ if $x\lt0$. Which parts of my answer do you fail to understand? $\endgroup$ – Did Dec 3 '14 at 8:31
  • $\begingroup$ ok, I get the + part now, but then you follow that line by "thus" and then u(x) and I cannot imagine how you went from one line to the next @Did $\endgroup$ – FrostyStraw Dec 3 '14 at 8:35
  • $\begingroup$ nevermind I see you just edited, so I'm gonna read our answer again @Did $\endgroup$ – FrostyStraw Dec 3 '14 at 8:35

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 $\lambda e^{-\lambda} y$).


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