I am a beginner in working with statistics and I wanted to generate a bivariate Poisson distribution $(X_1, X_2)$ and to do so, I did according to the indication found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution), therefore, I have generated three independent random variables which have a Poisson distribution $(Y_1, Y_2, Y_3)$, with means $\lambda_1 = \lambda_2 = \lambda_3 = 2$.

Provided I have set:

$X_1 = Y_1 + Y_3$

$X_2 = Y_2 + Y_3$


$X_1$ ~ Poisson($2+2$) = Poisson($4$)

$X_2$ ~ Poisson($2+2$) = Poisson($4$)

In order to compute the bivariate probability I use the formula found on wikipedia (bottom of the page, http://en.wikipedia.org/wiki/Poisson_distribution). When I try to compute the following probabilities (both in Matlab - set the format short - and on the paper), I obtain:

$P(X_1 = 4, X_2 = 4) = 0.0446$

$P(X_1 = 3, X_2 = 3) = 0.0474$

Basically, it indicates that $P(X_1 = 4, X_2 = 4) < P (X_1 = 3, X_2 = 3)$, which seems completely counterintuitive, since the mean for both random variables $X_1$ and $X_2$ is $4$. Initially, I have doubted my Matlab code and my calculation skills, but I tested it thouroughly and I obtain the same results. Is there any potential explanation for this or I am simply wrong in performing my calculations?

Thank you all.

  • 2
    $\begingroup$ The mean of a distribution does not have to be equal to its mode. $\endgroup$
    – Henry
    May 5, 2015 at 11:21
  • 1
    $\begingroup$ Your calculations look correct: $Pr(X_1=4,X_2=4)=\dfrac{156}{9}\exp({-6})$ while $Pr(X_1=3,X_2=3)=\dfrac{172}{9}\exp({-6})$ $\endgroup$
    – Henry
    May 5, 2015 at 11:44

2 Answers 2


No, you are right. I cannot do anything else but to reinforce your results based on the following calculations:

On the one hand $$P(X_1=4 \cap X_2=4)=P(Y_1+Y_3=4 \cap Y_2+Y_3=4)=\sum_{i=0}^4P(Y_1+Y_3=4 \cap Y_2+Y_3=4 \cap Y_3=i)=$$ $$=\sum_{i=0}^4P(Y_1=4-i)^2P(Y_3=i)=$$ $$=P(Y_1=4)^2P(Y_3=0)+P(Y_1=3)^2P(Y_3=1)+P(Y_1=2)^2P(Y_3=2)$$ $$+P(Y_1=1)^2P(Y_3=3)+P(Y_1=0)^2P(Y_3=4)=0.044617... .$$ On the other hand

$$P(X_1=3 \cap X_2=3)=P(Y_1+Y_3=3 \cap Y_2+Y_3=3)=\sum_{i=0}^3P(Y_1+Y_3=3 \cap Y_2+Y_3=3 \cap Y_3=i)=$$ $$=\sum_{i=0}^3P(Y_1=3-i)^2P(Y_3=i)=$$ $$=P(Y_1=3)^2P(Y_3=0)+P(Y_1=2)^2P(Y_3=1)$$ $$+P(Y_1=1)^2P(Y_3=2)+P(Y_1=0)^2P(Y_3=3)=0,047371... $$


The probability mass of a PoissonDistribution[4] is greater at $x = 3$ than $x = 4$, as you can easily calculate.

  • 2
    $\begingroup$ I suspect they are the same $\endgroup$
    – Henry
    May 5, 2015 at 11:23
  • $\begingroup$ Suspect?!!! In mathematics you suspect?!! My oh my! $\endgroup$ May 5, 2015 at 11:25
  • 2
    $\begingroup$ I was trying to be polite, inviting you to check your statement. They are both $\frac{32}{3} \exp(-4)$ $\endgroup$
    – Henry
    May 5, 2015 at 11:29
  • $\begingroup$ OK. Fair enough. Thanks. $\endgroup$ May 5, 2015 at 11:34
  • $\begingroup$ According to Matlab they are the same, as Henry pointed out :). However, in my case I am talking about a bivariate distribution, hence, I am observing the behavior of both variables at the same time, and my uncertainty came from the fact that the result I have obtained seemed counterintuitive, so I just assumed I am missing something in my reasoning :). $\endgroup$
    – SunnyDay
    May 5, 2015 at 11:39

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