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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 | ||
$${f}\left({x}\right)\:=\:\underset{{i}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{{i}} \frac{{x}^{\mathrm{2}{i}} }{{i}!}\:=\:{e}^{−{x}^{\mathrm{2}} } \\ $$ $$\int_{{x}>\mathrm{0}} {f}\left({x}\right){dx}\:=\:\int_{\mathrm{0}} ^{+\infty} {e}^{−{x}^{\mathrm{2}} } {dx}\:=\:\frac{\sqrt{\pi}}{\mathrm{2}}\:\neq\:\mathrm{1} \\ $$ $$\mathrm{This}\:\mathrm{is}\:\mathrm{not}\:\mathrm{a}\:\mathrm{pdf}. \\ $$ | ||