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

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

Answered by Olaf_Thorendsen last updated on 29/Jul/21

$$f\left(x\right)=\underset{i=0}{\sum ^{\mathrm{\infty }}}{(-1)}^i\frac{x^{2i}}{i!}=e^{-x^2}$$ $${\int }_{x>0}f\left(x\right)dx={\int }_0^{+\mathrm{\infty }}{e}^{-x^2}dx=\frac{\sqrt{\pi }}{2}\ne 1$$ $$\mathrm{This}\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{pdf}.$$

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