If both random variables X and Y are Poisson then their mean is equal to their variance respectively. I thought of subtracting both means but I realise, how was I going to get the variance. Poisson distribution=(μ^x.e^-μ)÷(x!). Where μ= mean, X can assume any number.
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2For independent random variables $X$, $Y$ and scalars $a,b\in\Bbb R$: $\mathsf{Var}(aX+bY)=a^2\mathsf{Var}X+b^2\mathsf{Var}Y$. – Aaron Hendrickson Jun 10 '21 at 22:39
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Thanks for the help, but how exactly can I solve for the mean . – Ernest Jun 10 '21 at 22:55
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Expectation is Linear (for any random variables). $\mathsf E(aX+bY)=a,\mathsf E,X+b,\mathsf E,Y$ – Graham Kemp Jun 10 '21 at 23:09
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Because $X,Y$ are Poisson,
$$E(X)=Var(X)=2\\ E(Y)=Var(Y)=5$$
Thus
$$E(X-Y)=2-5=-3\\ Var(X-Y)=2+5=7$$
Vons
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Thanks for your support, but please why is the variance 7 instead of 3. Then the mean will be equal to the variance making the distribution Poisson, can you please explain further. – Ernest Jun 10 '21 at 23:15
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@Ernest $Var(X-Y)=Var(X)+(-1)^2Var(Y)=2+5=7$. The coefficient of Y is squared, so it becomes +1 by a squaring operation. – Vons Jun 10 '21 at 23:17