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Question Number 148620 by jlewis last updated on 29/Jul/21

consider the following pdf of  a random variable X  f(x)={Σ_(i=0) ^∞ [(−x^2 )i/i!]_(0 otherwise) ^(x>0)    find the variance X

$$\mathrm{consider}\:\mathrm{the}\:\mathrm{following}\:\mathrm{pdf}\:\mathrm{of} \\ $$ $$\mathrm{a}\:\mathrm{random}\:\mathrm{variable}\:\mathrm{X} \\ $$ $$\mathrm{f}\left(\mathrm{x}\right)=\left\{\sum_{\mathrm{i}=\mathrm{0}} ^{\infty} \left[\left(−\mathrm{x}^{\mathrm{2}} \right)\mathrm{i}/\mathrm{i}!\right]_{\mathrm{0}\:\mathrm{otherwise}} ^{\mathrm{x}>\mathrm{0}} \:\right. \\ $$ $$\mathrm{find}\:\mathrm{the}\:\mathrm{variance}\:\mathrm{X} \\ $$ $$ \\ $$

Answered by Olaf_Thorendsen last updated on 29/Jul/21

f(x) = Σ_(i=0) ^∞ (−1)^i (x^(2i) /(i!)) = e^(−x^2 )   ∫_(x>0) f(x)dx = ∫_0 ^(+∞) e^(−x^2 ) dx = ((√π)/2) ≠ 1  This is not a pdf.

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

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