# Cumulative Distribution function of a Poisson distribution in terms of its parameter.

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 $$\lambda$$?

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

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 $$\sum_{k=n+1}^{\infty} p_k$$.

• Where is $dx$ in the last integral?
– Nick
Commented Sep 18, 2019 at 15:18
• 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 $x$: $\int_a^b x^2 \, dx = (b^3-a^3)/3$. Commented Sep 18, 2019 at 15:30
• I follow till $\frac {d}{d\lambda} P(X /le {n} ) = - p_n$ . How do you proceed with the next step? Commented Sep 18, 2019 at 15:47
• $\int_a^b f'(x) \, dx = f(b)-f(a)$, 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 $\lambda$. Commented Sep 18, 2019 at 15:52
• In the telescoping sum, for k = 1, where did $p_0(\lambda)$ go? That’s non zero right? 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.