Question Number 148620 by jlewis last updated on 29/Jul/21
$$\mathrm{consider}\:\mathrm{the}\:\mathrm{following}\:\mathrm{pdf}\:\mathrm{of} \\ $$$$\mathrm{a}\:\mathrm{random}\:\mathrm{variable}\:\mathrm{X} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\left\{\sum_{\mathrm{i}=\mathrm{0}} ^{\infty} \left[\left(−\mathrm{x}^{\mathrm{2}} \right)\mathrm{i}/\mathrm{i}!\right]_{\mathrm{0}\:\mathrm{otherwise}} ^{\mathrm{x}>\mathrm{0}} \:\right. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{variance}\:\mathrm{X} \\ $$$$ \\ $$
Answered by Olaf_Thorendsen last updated on 29/Jul/21
$${f}\left({x}\right)\:=\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{{x}^{\mathrm{2}{i}} }{{i}!}\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$$$\int_{{x}>\mathrm{0}} {f}\left({x}\right){dx}\:=\:\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\neq\:\mathrm{1} \\ $$$$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{pdf}. \\ $$