Clément Canonne @ccanonne.github.io · Apr 28 Let's look again at X~Poisson(λ). We have its MGF: let's Taylor expand it at 0, to order 4. (With some practice, it's easy and systematic, and anyways, Wolfram Alpha exists.) That's all, only need to read the coef. 𝔼[X⁴]=λ(λ³+6λ²+7λ+1). Want 𝔼[X³], 𝔼[X⁵] too? No problem, Taylor expand a bit more!
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Now, you may say: "but you cheated, you used Wolfram Alpha!" Yes, but frankly, Taylor expanding is not hard, it's only a few lines. e^t-1 = t+t²/2+t³/6+t⁴/24 + O(t⁵) so exp(λ(e^t-1)) = exp(λ(t+t²/2+t³/6+t⁴/24 + O(t⁵)))=1+λ(t+t²/2+t³/6+t⁴/24 + O(t⁵))+λ²(t+t²/2+t³/6+t⁴/24 + O(t⁵))²/2+..., etc. And..