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View Full Version : Problem 23.6 (expected value of a continuous random variable)

eagle121203
February 21st 2010, 10:45 PM
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
February 22nd 2010, 12:47 AM
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
February 22nd 2010, 01:51 AM
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
February 22nd 2010, 03:43 PM
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
February 22nd 2010, 05:37 PM
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
March 2nd 2010, 03:47 PM
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
March 15th 2010, 01:56 AM
hey guys, just use integration by parts. let u=lnx, and then du=1/x dx

NoMoreExams
March 15th 2010, 02:00 AM
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
March 16th 2010, 03:06 PM
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 :)