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If I take only the integer part of a Gaussian distribution, do I obtain a Poisson distribution?

In my python code, I created some integer values following the Gaussian distribution. Are these values following a Poisson distribution?

My python code: R = int(random.gauss(mu, sigma))

Moreover, in my code, the value R has to be positive: if R < 1: R = 1

Do these values still follow a Poisson distribution or this is a different distribution?

This is an example of my distribution with mean(mu)=10 and sigma=5 This is an example of my distribution with mean(mu)=10 and sigma=5

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No, that is not a Poisson distribution. You could call it a "discretized truncated normal distribution" if you want to, but that is not standard nomenclature.

Here are two ways to see that this is in general not a Poisson:

  • It will not give you zeros. Any Poisson distribution has probability mass at zero.
  • The Poisson distribution's variance is equal to its mean. Your distribution has two parameters and can thus have a variance that differs from the mean.

It may be possible to approximate a $\text{Pois}(\lambda)$ distribution by setting $\mu$ and $\sigma$ appropriately, but the discretization in particular will make this a complicated thing. It's much easier to deal with the Poisson directly, which is a well understood distribution.

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  • $\begingroup$ Many thanks, I was a bit lost. I will keep using my distribution and I will call it "discretized truncated normal distribution" as you suggested. This because I want to keep using both the parameters μ and σ. P.S. I put an image of my distribution in the main question. $\endgroup$ Jun 9 at 8:34
  • $\begingroup$ The specific distribution you plot is not Poisson for a third reason: the Poisson distribution is unimodal, but yours has two modes (at 1 and 8). It is more similar to a shifted zero-inflated Poisson (where the shift moves everything one bin to the right). $\endgroup$ Jun 9 at 8:39
  • $\begingroup$ If $\lambda$ is very large it should be possible to approximate by normal, but not a good approximation for small $\lambda$. $\endgroup$
    – Ben
    Jun 25 at 7:47
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I'll just consider the case with a standard Gaussian. Let $X \sim \mathcal N(0,1)$ and let $Y = \lfloor X\rfloor^+$, where this notation means to take the floor of $X$ and set it to zero if it's negative. You limited your support to $\{1,2,\dots\}$ but I'm changing the support to include zero so that the support of $Y$ is at least the same as a Poisson (as Stephan Kolassa notes in his answer, $Y$ would immediately be ruled out from being Poisson if this was not done).

I'm going to prove that $Y$ is not Poisson. I'll assume it is and then derive a contradiction.

$\newcommand{\Pr}{P_\text{real}}\newcommand{\Pp}{P_\text{Pois}}$First, I'll work out the actual distribution of $Y$. I'll use $\Pr$ for the actual distribution and $\Pp$ for the distribution under the assumption that $Y \sim \text{Pois}(\lambda)$.

For $y\geq 0$ we have $$ \Pr(Y = y) = \begin{cases}P(X < 1) & y = 0 \\ P(y \leq X < y+1) & y > 0\end{cases} \\ = \begin{cases}\Phi(1) & y = 0 \\ \Phi(y+1) - \Phi(y) & y > 0\end{cases} $$ where $\Phi$ is the standard Gaussian CDF.

By assumption I have $\Pr(Y=y) = \Pp(Y=y)$ for every $y$, so in particular I must have $$ \Pp(Y=0) = e^{-\lambda} = \Pr(Y=0) = \Phi(1) \\ \implies \lambda = -\log \Phi(1). $$ The Poisson distribution has one degree of freedom in $\lambda$ and this relationship fixes it uniquely. Now I want to show that $\Pr$ and $\Pp$ disagree on other values, giving my contradiction.

I tried $y=1$ which leads me to $$ \Pr(Y=1) = \Phi(2)- \Phi(1) \stackrel ?= -\Phi(1)\log\Phi(1) = \Pp(Y=1). $$ Numerically I can see that these are not equal, but that's not a proof. They're also fairly close so my intuition is that it will be challenging to bound one away from the other.

So next I tried the limit: if I can show that $$ \lim_{y\to\infty} \frac{\Pr(Y=y)}{\Pp(Y=y)} \neq 1 $$ then eventually they must disagree. Numerically it appears that this limit in fact is zero, which is good news for my proof, so now I'll try to show that.

Claim: $$ \lim_{y\to\infty} \frac{(2\pi)^{-1/2}\int_y^{y+1}e^{-x^2/2}\,\text dx }{e^{-\lambda}\lambda^y / y!} = 0. $$

Pf: I'll start with trying to show $$ \lim_{y\to\infty} \frac{a_y}{b_y} = 0 $$ where $a_y = \int_{y}^{y+1}e^{-x^2/2}\,\text dx$ and $b_y = \lambda^y / y!$, since then the results holds when I multiply back in the constants.

Since $e^{-x^2/2}$ is strictly decreasing on $[y,y+1]$ for large $y$, I can bound $a_y$ from above with $a'_y = e^{-y^2/2}$. If $a'_y/b_y\to 0$ then $a_y/b_y\to 0$ as well.

I'll now consider $$ L = \log \lim_{y\to\infty} \frac{a'_y}{b_y} = \lim_{y\to\infty} -\frac{y^2}2 + y\log \lambda^{-1} + \log y!. $$

I'm writing it as $\log \lambda^{-1}$ because $\lambda < 1$ (this is because $\lambda = -\log \Phi(1)$ and $\frac 12 < \Phi(1) < 1$ so it's a double negative) so this way I can see that I need to effectively show that $y + \log y!$ is outpaced by $y^2$.

$\log y! = \sum_{k=1}^y \log k < y \log y$ so I have $$ -\frac {y^2}2 + y \log \lambda^{-1} + \log y! < -\frac {y^2}2 + y \log \lambda^{-1} + y\log y \to -\infty $$

so this means $L = -\infty$, implying the original limit is zero.

$\square$

This proves that $Y$ cannot be Poisson.

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    $\begingroup$ +1 for a good idea. Perhaps a simpler demonstration is afforded by recognizing the Poisson survival function is given by the tail of a Gamma distribution. One application of L'Hospital's rule to their ratio does the trick. $\endgroup$
    – whuber
    Jun 25 at 13:46
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    $\begingroup$ @whuber ah thank you, I was trying CDFs and PMFs but hadn't thought of survival functions $\endgroup$
    – jld
    Jun 25 at 15:27

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