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Question Number 148864 by jlewis last updated on 31/Jul/21

$$\mathrm{A}\mathrm{random}\mathrm{variable}\mathrm{K}\mathrm{has}\mathrm{a}\mathrm{pdf}$$ $$\mathrm{f}\left(\mathrm{k}\right)=\{{}_{0,\mathrm{otherwise}}^{{\mathrm{e}}^{-\mathrm{k},0>\mathrm{k}}}\mathrm{find}\mathrm{E}\left(\mathrm{K}\right)\mathrm{and}$$ $$\mathrm{the}\mathrm{cdf}$$

Answered by Olaf_Thorendsen last updated on 01/Aug/21

$$\mathrm{I}\mathrm{suppose}f\left(k\right)=e^{-k}\mathrm{if}k>0\mathrm{otherwise}$$ $$f\mathrm{is}\mathrm{not}\mathrm{a}\mathrm{pdf}.$$ $${\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}f\left(x\right)dx={\int }_0^{+\mathrm{\infty }}{e}^{-x}dx=1$$ $$\mathrm{E}\left(\mathrm{K}\right)={\int }_0^{+\mathrm{\infty }}{xe}^{-x}dx={\left[x(-e^{-x})\right]}_0^{+\mathrm{\infty }}$$ $$-{\int }_0^{+\mathrm{\infty }}(-e^{-x})dx=1$$ $$\mathrm{E}\left(\mathrm{K}\right)=1$$ $$\mathrm{F}\left(k\right)={\int }_0^k{e}^{-x}dx=1-e^{-k}$$ $$\mathrm{F}\left(k\right)=\{\begin{array}{l}0\mathrm{if}k<0\\ 1-e^{-k}\mathrm{if}k⩾0\end{array}$$

Commented byjlewis last updated on 01/Aug/21

$$\mathrm{hello}\mathrm{sir},\mathrm{from}{\left[\mathrm{x}(-{\mathrm{e}}^{-\mathrm{x}})\right]}_0^{+\mathrm{\infty }}\mathrm{I}\mathrm{dont}\mathrm{get}$$ $$\mathrm{where}\mathrm{x}\mathrm{has}\mathrm{gone},\mathrm{and}\mathrm{is}\mathrm{x}\mathrm{a}\mathrm{variable}?$$

Commented byOlaf_Thorendsen last updated on 01/Aug/21

$$\mathrm{I}\mathrm{noted}\mathrm{the}\mathrm{variable}x,\mathrm{but}\mathrm{you}\mathrm{can}$$ $$\mathrm{write}\mathrm{instead}\mathrm{of}x:\mathrm{z},\mathrm{flower},\mathrm{Mickey}\mathrm{or}$$ $$\mathrm{Donald}\mathrm{Trump}\mathrm{sir},\mathrm{if}\mathrm{you}\mathrm{prefer}.$$

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