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Random variable X has probability density function fx(x) = {cxe^-x/2 x > = 0, 0 otherwise. Sketch the PDF and find the following: a. The constant c b. The CDF Fx(x) |
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Answer» fx(x) \(=\begin{cases}cx\,e^{-\frac x2},&x\geq0\\0,&x<0\end{cases}\) ∵ \(_0\int^\infty\) fx(x) dx = 1 ⇒ \(_0\int^\infty\,cx\,e^{-x/2}dx=1\) ⇒ \(c\left[-2x\,e^{-x/2}-\int-2e^{-x/2}dx\right]^\infty_0=1\) ⇒ \(c\left[-2x\,e^{-x/2}-4e^{-x/2}\right]^\infty_0=1\) ⇒ 4C = 1 ⇒ C = 1/4 CDF = Fx(x) \(=\,_0\int^x\,f_T(t)\,dt\) \(=\,_0\int^x\,t/4\,e^{-t/2}df\) \(=1/4[-2\,te^{-t/2}-\int-2e^{-t/2}df]^x_0\) \(=1/4[-2t\,e^{-t/2}-4e^{-t/2}]^x_0\) \(=1/4[-2x\,e^{-x/2}-4e^{-x/2}+4]\) Hence, CDF of probability function is Fx(x) = -x/2 e-x/2 - e-x/2 + 1 |
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