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kirill115 [55]
4 years ago
9

(1 pt) let x be an exponential random variable with parameter \lambda = 4, and let y be the random variable defined by y = 2 e^x

. compute the probability density function of y:
Mathematics
1 answer:
kupik [55]4 years ago
4 0
Y=2e^X\iff X=\ln\dfrac Y2

Note that x\in[0,\infty), so that y\in[2,\infty).

F_Y(y)=\mathbb P(Y\le y)=\mathbb P(2e^X\le y)=\mathbb P\left(X\le\ln\dfrac y2\right)=F_X\left(\ln\dfrac y2\right)

F_X(x)=1-e^{-4x}
\implies F_Y(y)=1-e^{-4\ln\frac y2}=1-\dfrac{16}{y^4}

\implies f_Y(y)=\dfrac{\mathrm dF_Y(y)}{\mathrm dy}=\dfrac{64}{y^5}
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