My idea is by doing a transformation, let $X' = X / \lambda$, then $X = X' \lambda$
Since $X \sim Poisson(\lambda)$, $\displaystyle P(X'=k) = P(X = \lambda k) = \displaystyle \frac{\lambda ^ {k \lambda}e^{-\lambda}}{(k\lambda)!}$
Is this correct?
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Sign up to join this communityMy idea is by doing a transformation, let $X' = X / \lambda$, then $X = X' \lambda$
Since $X \sim Poisson(\lambda)$, $\displaystyle P(X'=k) = P(X = \lambda k) = \displaystyle \frac{\lambda ^ {k \lambda}e^{-\lambda}}{(k\lambda)!}$
Is this correct?
Comment. Not quite: You need to specify precisely what values the random variable $X^\prime$ can take: Consider an example in R with $\lambda = 5.$
set.seed(213)
x = rpois(10^5, 5)
table(x)
x
0 1 2 3 4 5 6 7 8
702 3393 8382 14181 17681 17424 14646 10365 6407
9 10 11 12 13 14 15 16 17
3612 1750 856 393 134 52 15 6 1
x.1 = x/5
summary(x.1)
Min. 1st Qu. Median Mean 3rd Qu. Max.
0.0000 0.6000 1.0000 0.9984 1.2000 3.4000
table(x.1)
x.1
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6
702 3393 8382 14181 17681 17424 14646 10365 6407
1.8 2 2.2 2.4 2.6 2.8 3 3.2 3.4
3612 1750 856 393 134 52 15 6 1