Problem 23.6 (expected value of a continuous random variable)

[eagle121203]

Let X be a continuous random variable with pdf

f(x)= (1/x) for 1< x < e

Find E(ln X).

I have forgotten the natural log stuff in my ten years since college. I know I really need to review this.

I am thinking it is integral from 1 to e of ln(x)(1/x)dx.

I might be setting it up incorrectly.

[eagle121203]

let u=ln x and du=(1/x)dx

This is the integral of udu, which is ((u^2)/2).

This leaves us with (ln x)^2 all divided by two evaluated from 1 to e.

This becomes 1/2 minus zero, which is one-half, just like Dr. Finan says.

If someone could let me know....correct or incorrect???

[brandond]

In the continuous case you know that E(g(x)), where g(x) is a function of x, is [integral]g(x)f(x)dx correct? So here yes, I think your reasoning is correct.

[williamwjh326]

Originally Posted by eagle121203

Let X be a continuous random variable with pdf

f(x)= (1/x) for 1< x < e

Find E(ln X).

I have forgotten the natural log stuff in my ten years since college. I know I really need to review this.

I am thinking it is integral from 1 to e of ln(x)(1/x)dx.

I might be setting it up incorrectly.

E(ln X)= integral lnx / x dx (1<x<e) = integral lnx d(lnx) (0<lnx<1) = integral y dy (0<y<1) = y^2/2 (0<y<1) = 1/2=.5

[williamwjh326]

Originally Posted by eagle121203

Let X be a continuous random variable with pdf

f(x)= (1/x) for 1< x < e

Find E(ln X).

I have forgotten the natural log stuff in my ten years since college. I know I really need to review this.

I am thinking it is integral from 1 to e of ln(x)(1/x)dx.

I might be setting it up incorrectly.

How did u do the 23.19?

[eagle121203]

William:

Do a search on keywords from 23.19 (like "system"). Someone else on here solved it better than I could....sorry to leave you hanging. I have been sick.

[Bballry1234]

hey guys, just use integration by parts. let u=lnx, and then du=1/x dx

[NoMoreExams]

Originally Posted by Bballry1234

hey guys, just use integration by parts. let u=lnx, and then du=1/x dx

Isn't that what the 2nd post says?

[Bballry1234]

Originally Posted by eagle121203

let u=ln x and du=(1/x)dx

This is the integral of udu, which is ((u^2)/2).

This leaves us with (ln x)^2 all divided by two evaluated from 1 to e.

This becomes 1/2 minus zero, which is one-half, just like Dr. Finan says.

If someone could let me know....correct or incorrect???

The latter part of his post made me want to confirm any uncertainty. hehe