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
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Expert's answer
2022-04-20T02:34:32-0400
a). The pdf of X is a density of X. The variable Y=logX has a normal distribution with parameters 10 and 4. Variable X>0, since it appears under the function log. Therefore, P(X≤0)=0. The distribution of X is similar to the log-normal distribution.
We have: P(logX≤z)=P(Y≤z)=∫−∞zg(y)dy, where g(y)=22π1e−21(2y−10)2. P(X≤s)=P(logX≤logs)=∫0log(s)g(y)dy. The derivative(due to the Leibniz’s Theorem) is: f(s)=(P(X≤s))′=g(log(s))sln(10)1=(22π)(ln(10))s1e−21(2log(s)−10)2.
Thus, the pdf of X is: f(s)=(22π)(ln(10))s1e−21(2log(s)−10)2
b). E[X]=∫0+∞(22π)(ln(10))1e−21(2log(s)−10)2ds. Make the change of variables: x=ln(s). Then, we receive: s=ex. E[X]=∫−∞+∞(22π)(ln(10))exe−21(2ln(10)x−10)2dx=e81(a2−100)∫−∞+∞2ln(10)2π1e−21(2ln(10)x−aln(10))2dx,
where a=4ln(10)+10. The last integral is equal to 1 (it becomes the Gaussian integral after simple changes). Thus, we get: E[X]=e21((2ln(10)+5)2−25).
Var[X]=E[X2]−(E[X])2.
E[X2]=∫0+∞(22π)(ln(10))se−21(2log(s)−10)2ds. Make the change s=ex. We get: E[X2]=∫−∞+∞(22π)(ln(10))e2xe−21(2ln(10)x−10)2dx=e81(b2−100)∫−∞+∞2ln(10)2π1e−21(2,ln(10)x−bln(10))2dx,
where b=8ln(10)+10. Thus, Var[X]=e21((4ln(10)+52)−25)−e((2ln(10)+5)2−25).
We received: E[X]=e21((2ln(10)+5)2−25), Var[X]=e21((4ln(10)+52)−25)−e((2ln(10)+5)2−25)
c). P(X≤1000)=∫01000(22π)(ln(10))s1e−21(2log(s)−10)2ds≈0.0002 (it is rounded to 4 decimal places)
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