Question #330802

The random variable Y = LogX has N(10,4) distribution. Find (a) The pdf of X (b) Mean and variance of X (c) P ( X ≤ 1000




1
Expert's answer
2022-04-20T02:34:32-0400

a). The pdf of XX is a density of XX. The variable Y=logXY=log\,X has a normal distribution with parameters 1010 and 44. Variable X>0X>0, since it appears under the function loglog. Therefore, P(X0)=0P(X\leq0)=0. The distribution of XX is similar to the log-normal distribution.

We have: P(logXz)=P(Yz)=zg(y)dy,P(log\,X\leq z)=P(Y\leq z)=\int_{-\infty}^zg(y)dy, where g(y)=122πe12(y102)2g(y)=\frac{1}{2\sqrt{2\pi}}e^{-\frac12\left(\frac{y-10}{2}\right)^2}. P(Xs)=P(logXlogs)=0log(s)g(y)dyP(X\leq s)=P(log\,X\leq log\,s)=\int_{0}^{log(s)}g(y)dy. The derivative(due to the Leibniz’s Theorem) is: f(s)=(P(Xs))=g(log(s))1sln(10)=1(22π)(ln(10))se12(log(s)102)2f(s)=(P(X\leq s))'=g(log(s))\frac{1}{s\,\,ln(10)}=\frac{1}{(2\sqrt{2\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-10}{2}\right)^2}.

Thus, the pdf of XX is: f(s)=1(22π)(ln(10))se12(log(s)102)2f(s)=\frac{1}{(2\sqrt{2\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-10}{2}\right)^2}

b). E[X]=0+1(22π)(ln(10))e12(log(s)102)2ds.E[X]=\int_{0}^{+\infty}\frac{1}{(2\sqrt{2\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{log(s)-10}{2}\right)^2}ds. Make the change of variables: x=ln(s)x=ln(s). Then, we receive: s=exs=e^x. E[X]=+ex(22π)(ln(10))e12(xln(10)102)2dx=e18(a2100)+12ln(10)2πe12(xaln(10)2ln(10))2dx,E[X]=\int_{-\infty}^{+\infty}\frac{e^x}{(2\sqrt{2\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{\frac{x}{ln(10)}-10}{2}\right)^2}dx=e^{\frac1{8}(a^2-100)}\int_{-\infty}^{+\infty}\frac{1}{2\,ln\,(10)\sqrt{2\pi}\,\,}e^{-\frac12\left({\frac{x-a\,ln(10)}{2\,ln(10)}}\right)^2}dx,

where a=4ln(10)+10a=4\,ln(10)+10. The last integral is equal to 11 (it becomes the Gaussian integral after simple changes). Thus, we get: E[X]=e12((2ln(10)+5)225)E[X]=e^{\frac1{2}((2\,ln(10)+5)^2-25)}.

Var[X]=E[X2](E[X])2Var[X]=E[X^2]-(E[X])^2.

E[X2]=0+s(22π)(ln(10))e12(log(s)102)2ds.E[X^2]=\int_{0}^{+\infty}\frac{s}{(2\sqrt{2\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{log(s)-10}{2}\right)^2}ds. Make the change s=exs=e^x. We get: E[X2]=+e2x(22π)(ln(10))e12(xln(10)102)2dx=e18(b2100)+12ln(10)2πe12(xbln(10)2,ln(10))2dx,E[X^2]=\int_{-\infty}^{+\infty}\frac{e^{2x}}{(2\sqrt{2\pi})\,\,(ln(10))}e^{-\frac12\left(\frac{\frac{x}{ln(10)}-10}{2}\right)^2}dx=e^{\frac1{8}(b^2-100)}\int_{-\infty}^{+\infty}\frac{1}{2\,ln\,(10)\sqrt{2\pi}\,\,}e^{-\frac12\left({\frac{x-b\,ln(10)}{2,ln(10)}}\right)^2}dx,

where b=8ln(10)+10.b=8\,ln(10)+10. Thus, Var[X]=e12((4ln(10)+52)25)e((2ln(10)+5)225)Var[X]=e^{\frac1{2}((4ln(10)+5^2)-25)}-e^{((2\,ln(10)+5)^2-25)}.

We received: E[X]=e12((2ln(10)+5)225)E[X]=e^{\frac1{2}((2\,ln(10)+5)^2-25)}, Var[X]=e12((4ln(10)+52)25)e((2ln(10)+5)225)Var[X]=e^{\frac1{2}((4ln(10)+5^2)-25)}-e^{((2\,ln(10)+5)^2-25)}

c). P(X1000)=010001(22π)(ln(10))se12(log(s)102)2ds0.0002P(X\leq1000)=\int_{0}^{1000}\frac{1}{(2\sqrt{2\pi})\,\,(ln(10))s}e^{-\frac12\left(\frac{log(s)-10}{2}\right)^2}ds\approx0.0002 (it is rounded to 4 decimal places)


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