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Assuming $X\sim\mathcal{P}(\lambda)$ and $Y\sim\Gamma(w,1)$ prove that $P(X\ge w)=P(Y\le \lambda)$. How this fact is lead from the connection between the poisson and exponential distributions?

I don't know from where to start. poisson is defined only for discrete situations but the exponential is only for continious situation. How can I prove the fact ?
EDIT: for gamma distribution I wrote that $f_Y(y)=\frac{y^{w-1}e^{-y}}{\Gamma(w)}$, but I have problem with integrating it. About Poisson: its function is $\displaystyle \sum _{w_i=0}^w P(X=w_i)$ which I don't know how to sum into a final expression. How can I continue?

Michael Hardy
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  • Is there an assumption that $w$ should be an integer? – Stefan Hansen Aug 14 '13 at 11:58
  • No. it can be any real number. –  Aug 14 '13 at 13:03
  • Not sure it's true when $w$ is not an integer. At least, I can't show it unless $w$ is assumed to be an integer. – Stefan Hansen Aug 14 '13 at 17:59
  • This is true for every positive real number $\lambda$ and every positive integer $w$. If $w$ is not an integer this is wrong. The proof relies on the fact that $Y$ can be realized as $Z_1+Z_2+\cdots+Z_w$ for some i.i.d. standard exponential random variables $(Z_k)_{1\leqslant k\leqslant w}$. – Did Aug 14 '13 at 20:45
  • Assumeing $w$ is an integer how can can I prove it? –  Aug 15 '13 at 00:37
  • Should be integrating things like $\frac{1}{2!}\int_0^{\lambda} x^2 e^{-x} dx$ (for $k=3$). By the way, $\Gamma(w) = (w-1)!$ in that formula. In fact, don't bother with the Poisson side, you'll see it when you integrate by parts (take derivative of the polynomial, antiderivative of the exponential). I was just saying that for instance, $P(X \geq 2) = 1 - P(X=1) - P(X=0)$ – Evan Aug 15 '13 at 02:32
  • How the last equalation proves the fact for k=3? –  Aug 15 '13 at 02:38
  • If each inter-arrival time has an Exponential distribution then the n-th arrival time would be the sum of the N arrival times. So all you need to prove is that the Gamma is the sum of Exponential variables. – Fraijo Aug 15 '13 at 03:11
  • https://math.stackexchange.com/q/2205444/321264 – StubbornAtom Sep 28 '20 at 03:45

4 Answers4

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Let $w$ be an integer. Then $$ P(X\geqslant w)=1-P(X\leqslant w-1)=1-e^{-\lambda}\sum_{k=0}^{w-1}\frac{\lambda^k}{k!}. $$ Now, we use that if $\Gamma(a,b)$ denotes the incomplete Gamma function, i.e. $$ \Gamma(a,b)=\int_b^\infty t^{a-1}e^{-t}\,\mathrm dt, $$ then $$ \Gamma(n,b)=(n-1)!e^{-b}\sum_{k=0}^{n-1}\frac{b^k}{k!} $$ provided that $n$ is an integer. Thus (recall that $\Gamma(n)=(n-1)!$ for $n$ integer) $$ P(X\geqslant w)=1-\frac{\Gamma(w,\lambda)}{\Gamma(w)}. $$ Rewriting this expression we arrive at the desired expression $$ \begin{align} P(X\geqslant w)&=\frac{\Gamma(w)-\Gamma(w,\lambda)}{\Gamma(w)}=\frac{1}{\Gamma(w)}\left(\int_0^\infty t^{w-1}e^{-t}\,\mathrm dt-\int_\lambda^\infty t^{w-1}e^{-t}\,\mathrm dt\right)\\ &=\frac{1}{\Gamma(w)}\int_0^\lambda t^{w-1}e^{-t}\,\mathrm dt=P(Y\leqslant\lambda). \end{align} $$

Stefan Hansen
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Assuming $X\sim\mathcal{P}(\lambda)$ and $Y\sim\Gamma(w,1)$ prove that $P(X\ge w)=P(Y\le \lambda)$.

Suppose the number of occurrences during a time interval of length $\lambda$ is a Poisson-distributed random variable with expected value $\lambda$, and the numbers of occurrences in disjoint time intervals are independent.

Then the number of occurrences before time $\lambda$ is zero if and only if the waiting time until the first occurrence is more than $\lambda$.

The probability that a $\operatorname{Poisson}(\lambda)$-distributed random variable is $0$ is $e^{-\lambda}$.

Therefore the probability that the waiting time until the first occurrence is more than $\lambda$ is $e^{-\lambda}$.

In other words, the waiting time has an exponential distribution with expected value $1$.

If you know that the sum of independent exponentially distributed random variables is Gamma-distributed, then you've got it.

Punch line: The number of occurrences before time $\lambda$ is at least $w$ if and only if the waiting time until the $w$th occurrence is less than $\lambda$. (And since it's a continuous distribution, that's the same as the probability that it's less than or equal to $\lambda$.)

Michael Hardy
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Oh right... this problem justifies the intuition that connects Poisson with Exponential.

If you want the raw way of doing it, which at this early stage might be the only option anyway, look at the pdf for the Gamma function and just compute directly the probability of being $\leq \lambda$. Repeated integration by parts reveals the Poisson distribution formula that you want.

Also should say, the interpretation of this formula is the beautiful part. If you imagine being a store owner and suppose customers arrive in your store, and the time you wait between arrivals is exponentially distributed with rate $1$. Then you can ask, what is the distribution of the number of customers that arrive within a time interval $\lambda$?

A sample calculation for $w=2$:

\begin{align} P(Y_2 \leq \lambda) &= \frac{1}{1!} \int_0^\lambda xe^{-x} \, dx\\ &= \left[xe^{-x}\right]_0^\lambda - \int_0^\lambda e^{-x} \, dx \\ &= 1 - \lambda e^{-\lambda} - e^{-\lambda} \\ &= P(X \geq 2) \end{align}

I suppose at some point with this approach you'd have to use induction.

Michael Hardy
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Evan
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  • You are using http://en.wikipedia.org/wiki/Poisson_distribution#Definition ? and http://en.wikipedia.org/wiki/Gamma_distribution#Probability_density_function ? (with $\theta=1$?) – Evan Aug 15 '13 at 01:59
  • Oh, and if you are trying to compute an infinite sum, you are forgetting something simpler (complements) – Evan Aug 15 '13 at 01:59
  • I used them both and still failed. Why the sum in poisson is infinite? –  Aug 15 '13 at 02:00
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    Oh, it's not infinite. I'm saying you should only be subtracting finitely many things for Poisson. As for gamma, did you at least work out $w=1$ and $w=2$ case? PS you might want to post partial work in your post too, to see what you tried. – Evan Aug 15 '13 at 02:01
  • I am still stuck with simplifying the definitions (see my edit). –  Aug 15 '13 at 02:14
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I don't find the currently accepted answer particularly satisfying since it seems almost circular in assuming a formula that is essentially the answer....

Just do a single integration by parts:

$\frac{1}{(w-1)!} \int_0^\lambda x^{w-1} e^{-x} dx = -\lambda^{w-1}e^{-\lambda}/(w-1)!+ \frac{1}{(k-2)!} \int_0^\lambda x^{w-2} e^{-x} dx$

Gives

$P(Y_w <= \lambda) = P(Y_{w-1} <= \lambda) - P(X = w-1)$

Use induction to finish.

Evan
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