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Let $X,Y$ be independent random variables with $X,Y \sim$ Poi$(\lambda)$. What is the distribution of $X-Y$? Since $X$ and $Y$ are independent I would assume that $X$ and $-Y$ are independent and therefore it follows that $$\varphi_{X-Y}(t) = \mathbb E[\exp(it(X-Y))] = E\left[\frac{\exp(it(X))}{\exp(it(Y))}\right] = \frac{\exp(\lambda(e^{it}-1))}{\exp(\lambda(e^{it}-1))} = 1.$$ But as far as I know $X-Y\sim$ Poi$(\lambda)$. Should I consider introducing a random variable $Z$ with $\mathbb P(Z = 1) = \mathbb P(Z=-1)=\frac 1 2$ with $\varphi_Z(t) = \cos(t)$?

Thank you!

HelloWorld
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  • You need to take another look at the characteristic function for $-Y$. The characteristic function for the difference is $e^{2 \lambda (\cos (t)-1)}=\exp \left(\lambda \left(-1+e^{-i t}\right)+\lambda \left(-1+e^{i t}\right)\right)$. – JimB Nov 20 '22 at 17:52
  • @JimB thank you. I think it should be correct now. – HelloWorld Nov 20 '22 at 19:51
  • @Snoop this is helpful. – HelloWorld Nov 20 '22 at 20:00

1 Answers1

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As it turns out the characteristic function is not 1 it is \begin{align*}\varphi_{X-Y}(t) &= \mathbb E[\exp(it(X-Y))]\\ &=\mathbb E[\exp(itX)]\mathbb E[\exp(-itY)] \\ &=\exp(\lambda(e^{it}-1))\exp(\lambda(e^{-it}-1))\\ &=\exp(2\lambda(e^{it}-e^{-it}-1))\\ &=\exp(2\lambda(\cos(t)-1)). \end{align*}

The corresponding distribution is not a Poisson distribution. As the mean of $X-Y$ is $\lambda - \lambda = 0$ and the variance $\lambda + \lambda = 2\lambda$. It is a Skellam distribution as Snoop pointed out. For $X,Y$ with $X\sim$ Poi$(\lambda_1), Y\sim$ Poi$(\lambda_2)$ the difference $X-Y$ would have mean $\lambda_1-\lambda_2$ and variance $\lambda_1+\lambda_2$.

HelloWorld
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