21.Let \(X\) be a random variable which is Poisson distributed with parameter \(\lambda\). Show that \(E(X)\)=\(\lambda\).
Hint: \(e^x\) = 1 + \(x\) + \(\frac{x^2}{2!}\) + \(\frac{x^3}{3!}\) + …
\(e^x\) = 1 + \(x\) + \(\frac{x^2}{2!}\) + \(\frac{x^3}{3!}\) + …
\(e^\lambda\) = \(\sum_{x=0}^{\infty}\frac{\lambda^x}{x!}\) = \(\sum_{x=1}^{\infty}\frac{\lambda^{x-1}}{(x-1)!}\)
\(E[X]\) = \(\sum_{x=0}^\infty x p(x)\)
= \(\sum_{x=0}^\infty x e^{-\lambda}\frac{\lambda ^x}{x!}\)
= \(e^{-\lambda}\sum_{x=0}^{\infty}x\frac{\lambda ^x}{x!}\)
= \(e^{-\lambda}\sum_{x=1}^{\infty}x\frac{\lambda ^x}{x!}\)
= \(e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda ^x}{x-1!}\)
= \(e^{-\lambda}\sum_{x=1}^{\infty}\frac{\lambda ^{x-1}\lambda}{x-1!}\)
= \(e^{-\lambda}\lambda\sum_{x=1}^{\infty}\frac{\lambda ^{x-1}}{(x-1)!}\)
= \(e^{-\lambda}\lambda e^\lambda\)
= \(\lambda (e^{-\lambda})(e^{\lambda})\)
= \(\lambda\)