Variance of a multiple of a Poisson distribution

Question

a) Say we have two independent distribution variables $X \sim Pois(\lambda)$ and $Y \sim Pois(\mu)$. We know that the sum of the two will be $$X+Y\sim Pois(\lambda+\mu),$$ and hence the variance will be $Var(X+Y)=\lambda+\mu$.

b) We also know that in general $Var(b\cdot X) = b^2\cdot Var(X)$: $$ Var(bX) = E(b^2X^2) - E^2(bX) = b^2E(X^2) - b^2E^2(X) = b^2 Var(X) $$

However, if I take the distribution $X + X$, I have two different results for the variance:

a) $X + X \sim Pois(\lambda + \lambda) = Pois(\alpha)$, with $\alpha = 2\lambda$, and hence $$Var(X+X) = \alpha = 2\lambda$$ b) $$Var(X+X)=Var(2X)=2^2Var(X)=4\lambda$$

What am I doing wrong? According to my exercise sheet, a) is correct, but then I don't see why b) wouldn't apply. Thanks for any hints!

Is $X$ independent of itself when you apply the result (a) ? — StubbornAtom, Jan 06 '18 at 16:47
According to the master solution of my exercise sheet it seems to be, but then again, it could be a mistake... I am honestly not sure... — Sean Bone, Jan 06 '18 at 16:49
Have a look at this post:https://math.stackexchange.com/questions/512755/independence-of-a-random-variable-x-from-itself. — StubbornAtom, Jan 06 '18 at 16:49

Sri-Amirthan Theivendran · Accepted Answer · 2018-01-06T16:56:50.100

1

$X$ and $X$ are not independent. Hence you cannot apply result (a). In fact, you have shown that $2X$ is not a poisson random variable as $Var(2X)\neq E(2X)$ as $\lambda>0$.

edited Jan 06 '18 at 16:56

answered Jan 06 '18 at 16:49

Sri-Amirthan Theivendran

31,832

I see. Hence there really seems to be an error in my textbook. Thank you very much! – Sean Bone Jan 06 '18 at 16:54

Variance of a multiple of a Poisson distribution

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