Let $P(X=n) = \lambda^ne^{-\lambda}/n!$ which is a Poisson distribution. I need to find the cumulative distribution function.
The first think that comes to my mind is summation across n from $-\infty$ to $\infty$. But the answer is in terms of integration of $\lambda$.

$P(X\le n) = 1-\int_{0}^{\lambda} (x^ne^{-x}/n!)dx$

How do we change that integration over x into integration over $\lambda$?

  • $\begingroup$ Where are you getting this answer from? $\endgroup$ Commented Sep 18, 2019 at 15:05
  • 2
    $\begingroup$ This is because the CDF of Poisson distribution is related to that of a Gamma distribution. Hence the incomplete gamma function. $\endgroup$ Commented Sep 18, 2019 at 15:18
  • $\begingroup$ Plenty of duplicates: math.stackexchange.com/questions/467341/… $\endgroup$ Commented Sep 18, 2019 at 16:44

2 Answers 2


The Poisson distribution takes values $n \in \{0,1,2,\dotsc\}$ with probability $P(X=n) = e^{-\lambda} \lambda^n / n! =: p_n(\lambda) $, so the cumulative distribution function is $$ P(X \leq n) = \sum_{k=0}^n e^{-\lambda} \frac{\lambda^k}{k!} = \sum_{k=0}^n p_k(\lambda) . $$ This sum has a rather weird property (which admittedly one would not spot unless one is familiar with particular differential equations): if we differentiate with respect to $\lambda$, we find that $$ p_k'(\lambda) = \frac{d}{d\lambda} e^{-\lambda} \frac{\lambda^k}{k!} = - e^{-\lambda} \frac{\lambda^k}{k!} + e^{-\lambda} \frac{\lambda^{k-1}}{(k-1)!} = -p_k + p_{k-1} , $$ and $p_0'(\lambda) = 0$. Thus, $$ \frac{d}{d\lambda} P(X \leq n) = \sum_{k=0}^n p_n'(\lambda) = 0 + \sum_{k=1}^n (p_{k-1}(\lambda)-p_k(\lambda)) . $$ If we write out the sum, we see that it telescopes, leaving only $-p_n$. Hence, by the Fundamental Theorem of Calculus, $$ P(X \leq n) = P(X \leq n)(\lambda=0) - \int_0^{\lambda} p_n(x) \, dx . $$ The first term is $1$ since a Poisson distribution with parameter $0$ takes the value $0$ with probability $1$, the second is the integral given in the answer.

Of course everyone is wondering why you would want to do this. I think the simplest reason is, if you want to estimate the probability of being larger than $n$, it is much easier to do this with the expression $\int_0^{\lambda} p_n $ than the infinite sum $\sum_{k=n+1}^{\infty} p_k$.

  • $\begingroup$ Where is $dx$ in the last integral? $\endgroup$
    – Nick
    Commented Sep 18, 2019 at 15:18
  • $\begingroup$ For functions of one variable that are written as a set of symbols, it's quite common to omit the integration variable: it's redundant. For example, $$\int_0^x \sin = 1-\cos{x}$$ is completely unambiguous. Similarly, if I define a "squaring" function by $s(x) = x^2$, $ \int_a^b s = (b^3-a^3)/3 $, but you need the $dx$ if the function is written as an explicit expression containing $x$: $\int_a^b x^2 \, dx = (b^3-a^3)/3$. $\endgroup$
    – Chappers
    Commented Sep 18, 2019 at 15:30
  • $\begingroup$ I follow till $\frac {d}{d\lambda} P(X /le {n} ) = - p_n$ . How do you proceed with the next step? $\endgroup$ Commented Sep 18, 2019 at 15:47
  • $\begingroup$ $ \int_a^b f'(x) \, dx = f(b)-f(a) $, so $ f(b) = f(a) + \int_a^b f'(x) \, dx $. In this case, $f$ is $P(X \leq n)$, regarded as a function of $\lambda$. $\endgroup$
    – Chappers
    Commented Sep 18, 2019 at 15:52
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    $\begingroup$ In the telescoping sum, for k = 1, where did $p_0(\lambda)$ go? That’s non zero right? $\endgroup$ Commented Jul 16, 2023 at 14:30

Yes, $p_0(\lambda)$ is there, so the telescope sum is $p_0-p_n$. Only it drops out in the final formula, as the derivative/integral of $p_0(\lambda)$ is $-p_0(\lambda)$, so the final formula is OK.


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