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Let $Y_1\sim \exp(\lambda_1)$ and $Y_2\sim \exp(\lambda_2)$ be two independent r.v.'s.

Show that the pdf $p_V(x)$ for their sum $V=Y_1+Y_2$ has the following form $$p_V(x)=\frac{e^\frac{-v}{\lambda_1}-e^\frac{-v}{\lambda_2}}{\lambda_1-\lambda_2};\quad v\ge0$$

My attempt:

distribution of $Y_i$, $$p_{Y_i}(y_i)=\frac{1}{\lambda_i}e^\frac{-{y_i}}{\lambda_i};\quad x\ge0;\quad i=1,2$$

joint distribution of $Y_1$ and $Y_2$, $$p_{Y_1,Y_2}(y_1,y_2)=\frac{1}{\lambda_1\lambda_2}e^{\frac{-{y_1}}{\lambda_1}-\frac{{y_2}}{\lambda_2}};\quad {y_1},{y_2}\ge0$$

Given, $V=Y_1+Y_2$

let $U=Y_2$

So the Jacobian of Transformation is $1$

and hence joint distribution of $V$ and $U$, $$p_{V,U}(v,u)=\frac{1}{\lambda_1\lambda_2}e^{\frac{-{v-u}}{\lambda_1}-\frac{{u}}{\lambda_2}};\quad {v},{u}\ge0$$

so the distribution of $v$ $$p_{V}(v)=\int_0^\infty p_{V,U}(v,u)du=\frac{e^\frac{-v}{\lambda_1}}{\lambda_1-\lambda_2};\quad v\ge0$$

which doesn't match with the result.

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ABC
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1 Answers1

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We could calculate the cumulative distribution function of $V$, and then differentiate. It is quicker to use the convolution $$\int_{-\infty}^\infty f_1(z-t)f_2(t)\,dt,$$ where $f_1$ and $f_2$ are the densities of our random variables.

The density function $f_V(v)$ of $V$ is $0$ for $v\lt 0$. So we look only at the case $v\ge 0$. In our case, since the exponentials have density $0$ to the left of $0$, the actual expression is $$f_V(v)=\int_{t=0}^v \lambda_1 \lambda_2 e^{-\lambda_1(v-t)}e^{-\lambda_2 t}\,dt.$$ We are basically integrating $e^{-(\lambda_2-\lambda_1)t}$, which is not difficult.

André Nicolas
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  • In one of my courses Sampling Distributions i read there are $4$ methods of finding sampling distributions.One of the methods are analytical method. The method can be used when there is only $2$ random variables.According to the method the distribution of the statistic $V=Y_1+Y_2$ is $$\int_{-\infty}^\infty f_1(z-t)f_2(t),dt,\ldots(1)$$. Could you please tell me (1)Why does not the range is $0$ to $\infty$ here? Why have you considered the cdf.If i consider eqn(1),this is pdf. – ABC Aug 24 '13 at 04:30
  • (2)Also in my attempt if i consider the range from $0$ to $v$, i come up with the result. I did many convolutions according to my attempt and analytical method with the range of statistic.There i faced no problem. – ABC Aug 24 '13 at 04:33
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    (1) I did not use the cdf. Just mentioned that it is an alternate way of finding the density. I actually prefer to do it that way when explaining things to students, since it is has a more conceptual feel. But I chose this time to use the convolution, because it is faster. (Continued) – André Nicolas Aug 24 '13 at 04:35
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    (2) We are integrating in principle from $-\infty$ to $\infty$, as I mentioned in the first formula. However, $\lambda_1 e^{-\lambda_1( v-t)}$ is only correct for the density of $Y_1$ when $v-t\ge 0$. Others like to use indicator functions to take care of this problem. It comes down to the same thing. – André Nicolas Aug 24 '13 at 04:41
  • I apologize. I haven't understood yet. If $v-t\ge0$ then $v\ge t$. Does it mean $v$ ranges $t$ to $\infty$? Let $X_i\sim^{iid}N(0,1);i=1,2$. When i tried to find the distribution of $Y=X_1+X_2$ by analytical method i found $$f(y)=\int^{-\infty}_{\infty}\frac{1}{\sqrt{2\pi}}e^\frac{-(y-x_2)^2}{2}\frac{1}{\sqrt{2\pi}}e^\frac{-{x_2}^2}{2}dx_2$$

    If i take the range from ${-\infty}$ to $y$, i don't come up with the result. Why will i take the range ${-\infty}$ to ${\infty}$ here?

     – ABC Aug 24 '13 at 05:00
  • For some reason, you are using the density function of the normal. Our random variables have exponential distribution. With the normal, there is no negativity issue, since the density function is given by a single formula everywhere. With the exponential, the density is $0$ at negative places. Also, $v$ is fixed for this calculation, we are integrating with respect to $t$. Also, $v-t\ge 0$, so $t$ can only range from $0$ to $v$. – André Nicolas Aug 24 '13 at 05:08
  • When the case is for $n$ independent random variables,how can i proceed ? – ABC Aug 25 '13 at 23:59
  • It is a mess unless the parameters are all equal. Then you get an Erlang distribution, can get to it by induction or (easier!) finding the $n$-th power of the moment generating function of the exponential, and checking against the mgf of the Erlang. – André Nicolas Aug 26 '13 at 00:10