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Can you please help me prove this identity below? I have tried integration by parts and I am able to see where the numerator of the RHS comes from; however, I am unable to get the denominator. Additionally, I am unsure of how to use induction to show this, because I know that I would have to use integration by parts $k-1$ times to prove this... thanks for the help!

$\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\sum_{x=0}^{k-1} \frac{(\alpha w)^x e^{-\alpha w}}{x!}$

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    No induction is needed here. Differentiate term by term the sum on the RHS with respect to $\alpha w$. Nearly everything cancels out and the net result is minus the function in the integral, evaluated at $z=\alpha w$. Since the LHS and the RHS go to 0 when $\alpha w\to+\infty$, the proof is complete.2011-09-25

4 Answers 4

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$\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\sum_{x=0}^{k-1} \frac{(\alpha w)^x e^{-\alpha w}}{x!}$

The right side is the probability that the number of occurrences in a Poisson process before time $w$ is less than $k$ (with an average of $\alpha$ occurrences per unit of time).

The left side is the probability that the waiting time until the $k$th occurrence is more than $w$.

One side looks at the discrete probability distribution of the number of occurrences in a given time interval; the other looks at the continuous probability distribution of the the time until the $k$th occurrence.

But the two events are the same event. So they must have the same probability. Therefore they are equal, quod erat demonstrandum.

Later edit: I find a complaint of cirucularity in the comments. That is mistaken. Find the probability distribution of the number of successes in $n$ independent Bernoulli trials with probability $p$ of success on each trial; take the limit as $n\to\infty$ while $np$ remains fixed at $\lambda$. The limiting probability of $x$ successes is $e^{-\lambda} \lambda^x/x!$. Say this is in unit time; the numbers of successes in non-overlapping time intervals are independent; then the number of successes before time $t$ has expected value $t\lambda$ and the probability of exactly $x$ successes is $e^{-t\lambda}(t\lambda)^x/x!$. Now the probability of fewer than $x$ successes before time $t$ is the sum of $x$ terms given above (but I'm using "$x$" where "$n$" was used). What then is the probability distribution of the waiting time until the $n$th success? The value of the cumulative distribution function at time $t$ is just $1$ minus the sum given above. Differentiate that to get the density. The integral of the density is the integral above.

So there's no circularity.

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    ....and of course integrating by parts, plus mathematical induction, will also do it!2011-09-26
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This one cries out for a generating function. For $|t|<1$, $\sum_{k=0}^\infty t^k \int_{\alpha w}^\infty \frac{z^{k}}{k!} e^{-z} \ dz = \int_{\alpha w}^\infty \sum_{k=0}^\infty \frac{(tz)^k}{k!} e^{-z} \ dz $ $ = \int_{\alpha w}^\infty e^{t z - z}\ dz = \frac{e^{(t-1) \alpha w}}{1-t} $ $ = e^{-\alpha w} \left(\sum_{k=0}^\infty \frac{(\alpha w t)^k}{k!}\right)\left(\sum_{k=0}^\infty t^k\right)$ $ = \sum_{k=0}^\infty t^k \sum_{j=0}^k \frac{(\alpha w)^j}{j!}$

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    My first comment may have been too concise, let me expand on it: usually, introducing generating functions provides a huge simplification of the problem, which I fail to see in the present case. // Unrelated: what happened to the exponential factor in your last step?2011-09-26
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Here's one way of proceeding: first, verify that both sides of the purported identity agree for $k=1$. Now, if you carry out integration by parts on the left hand side, you should obtain the recursion relation

$\beta_{k-1}=\beta_{k-2}+\frac{(\alpha w)^{k-1}e^{-\alpha w}}{(k-1)!}$

where

$\beta_{k-1}=\int_{\alpha w}^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=\frac{\Gamma(k,\alpha w)}{\Gamma(k)}=Q(k,\alpha w)$

and $\Gamma(k,u)$ and $Q(k,u)$ are various notations for the (upper) incomplete gamma function.

The trick now is to replace $\beta_{k-1}$ in the recursion relation with the sum on the right hand side, and verify that the recursion still holds. This along with the verified initial condition $k=1$ proves your identity.


As a variation of Didier's strategy in the comments: after letting $\alpha w=0$, to verify the equation

$\int_0^\infty \frac{z^{k-1}e^{-z}}{(k-1)!}\mathrm dz=e^{0}\left(1+\sum_{x=1}^{k-1} \frac{0^x}{x!}\right)$

one can demonstrate that the expression on the right does simplify to $1$; showing that

$(k-1)!=\int_0^\infty z^{k-1}e^{-z}\mathrm dz=\Gamma(k)$

can be done by verifying that both sides of the equation agree when $k=1$ and establishing the recursion $(k-1)!=(k-1)(k-2)!$ through integration by parts.

Differentiating both sides of the original identity with respect to $u=\alpha w$ yields the relation

$-\frac{u^{k-1}e^{-u}}{(k-1)!}=e^{-u}\left(\sum_{x=1}^{k-1} \frac{xu^{x-1}}{x!}-\sum_{x=0}^{k-1} \frac{u^x}{x!}\right)$

I'll leave the simplification to the reader.

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Another answer, my second one to this question:

Differentiate both sides with respect to $\omega$. On the left you need the chain rule in a trivial way and the fundamental theorem of calculus, and you get only one term. On the right you get $2k-1$ terms (after apply the product rule to all but one of the $k$ terms) and $2k-2$ of them cancel out. The highest-degree term is equal to the expression you get on the right side. That proves they differ by a constant; finding the constant is the remaining task.

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    Interesting solution. I am sure I saw it quite recently, but cannot seem to remember where.2011-09-26