3
$\begingroup$

I need to calculate this integral

$$\sum\limits_{k=1}^\infty\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k$$

where $I=(a,b)$.

Someone told me that summation equals $\lambda$, but I don't see why. I wrote the summation as

$$\sum\limits_{n=0}^\infty \frac{(\lambda t_{n+1})^n}{n!}\lambda e^{-\lambda t_{n+1}} $$

and I was trying to see if the fact that $\sum\limits_{n=0}^\infty \frac{(\lambda t)^n}{n!}=e^{\lambda t} $ could help but I'm not seeing how.

Any ideas?

$\endgroup$
  • 2
    $\begingroup$ Does $t_n$ depend on $n$? $\endgroup$ – Ahmed S. Attaalla Jul 1 '16 at 20:17
  • $\begingroup$ @AhmedS.Attaalla Yes, $T_n$ are the arrival times of a Poisson process so they are different from each other and random variables each. The $t_n$ in the summation above appeared because I was trying to get the expectation of an expression involving those $T_n$, so I had to integrate its probability density function. I'm editing my question to clarify this. $\endgroup$ – Tendero Jul 1 '16 at 20:20
  • $\begingroup$ It looks me odd that $I=(t_1,t_2)$, $t_1$ and $t_2$ being the two first integration variables $t_k$, but may be I am wrong. $\endgroup$ – Jean Marie Jul 1 '16 at 20:33
  • $\begingroup$ @JeanMarie Yes that was an unfortunate naming, I just edited it. Thanks $\endgroup$ – Tendero Jul 1 '16 at 20:34
5
$\begingroup$

The confusion arises because of the bad choice of the name of the integration variable(s), as if it actually depended on $k$. Write the first few terms explicitly: $$ \sum\limits_{k=1}^\infty\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\int_I \frac{\lambda}{0!}t_1^{0}e^{-\lambda t_1}dt_1+ \int_I \frac{\lambda^2}{1!}t_2 e^{-\lambda t_2}dt_2+\cdots $$ You can easily see that calling the first integration variable $t_1$ and the second $t_2$ is unnecessary. Rename all the $t_k$'s as $t$. Your object is just a sum of one-dimensional integrals. Then $$ \sum_{k=1}^\infty\frac{\lambda^k}{(k-1)!}t^{k-1}=\lambda e^{t\lambda}\ . $$ Therefore the final result is $$ \lambda\int_a^b dt\ e^{-\lambda t}e^{t\lambda}=\lambda(b-a). $$ Note that whoever told you that the solution is just $\lambda$ ignores the fact that it is very unlikely that $a$ and $b$ (the integration range) drop out of the game!

$\endgroup$
3
$\begingroup$

There's no need for having different variables for integration. So we aim to solve $$ \sum_{k=1}^\infty\int_I\frac{\lambda^k}{(k-1)!}t^{k-1}e^{-\lambda t}\ dt =\int_I\sum_{k=0}^\infty\frac{\lambda^{k+1}}{k!}t^ke^{-\lambda t}\ dt$$ Pulling out the parts which are unnecessary for summation, we have \begin{align*} \sum_{k=1}^\infty\int_I\frac{\lambda^k}{(k-1)!}t^{k-1}e^{-\lambda t}\ dt &=\int_I\lambda e^{-\lambda t}\sum_{k=0}^\infty\frac{(\lambda t)^k}{k!}\ dt \\ &=\int_I\lambda e^{-\lambda t}e^{\lambda t}\ dt \\ &=\lambda(b-a) \end{align*}

$\endgroup$
1
$\begingroup$

You cannot a priori write the summation as you have done, because each time it's different variables. But you can do it a posteriori as I will show, using an appropriate change of variable which will "neutralize" the differences between the different $t_k$s. Here is how:

Let $u:=\lambda t_k$ and $J=(\lambda a, \lambda b)$:

$$\int\limits_I \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\frac{1}{(k-1)!}\int\limits_J u^{k-1}e^{-u} du$$

Thus, setting $A:=\int\limits_J u^{k-1}e^{-u} du$

$$\sum\limits_{k=1}^\infty\int\limits_J \frac{\lambda^k}{(k-1)!}t_k^{k-1}e^{-\lambda t_k} dt_k=\int\limits_J \sum\limits_{k=1}^\infty\frac{u^{k-1}}{(k-1)!}e^{-u} du=\int\limits_J e^{u}e^{-u} du=\int\limits_J 1 du=\lambda b- \lambda a=\lambda(b-a)$$

$\endgroup$
  • 1
    $\begingroup$ Isn't your $A$ depending on $k$? Why can you pull it out of the summation over $k$? $\endgroup$ – Pierpaolo Vivo Jul 1 '16 at 20:59
  • $\begingroup$ You are write : I have done a big mistake... I correct it... $\endgroup$ – Jean Marie Jul 1 '16 at 21:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.