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so I'm learning a new topic and am a bit new to probability, so please excuse my elementary question.

Given a random sample on an exponential distribution with mean $\theta$ of $X_1,X_2,...,X_n$, let $Y = \ln{X}$. Find the mean of $Y$.

We are also given the gamma function, $\Gamma(z) = \int_0^\infty t^{z-1}e^{-t}dt$

My approach is to first find the CDF of $Y$, then it's PDF, and then calculating expectation via $E[Y] = \int y \cdot f_Y(y)dy$

Would this yield the correct answer?

Is there a faster, more efficeint approach?

Have I made a mistake anywhere?

Thanks in advance for your help.

Approach so far: (Has been added at the end of comments)

How can I go from using a given result that

$$\gamma = - \int_0 ^\infty ln(x) e^{-x} dx = 0.5772..$$ When I currently have :

$$E[\ln (X) ] = \frac{1}{\theta}\int_0 ^\infty ln(x) e^{\frac{-x}{\theta}} dx$$

LearningMath
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    What is the random variable $X$? In general the procedure you outlined is fine, but there may be simpler ways to find the mean. – André Nicolas Sep 03 '15 at 16:52
  • $X$ is exponentially distributed with mean $\theta$ - sorry for not making it clear! I will update my post with an attempt at the solution in a bit, hopefully someone can check :) – LearningMath Sep 03 '15 at 16:58
  • I thought it might be the the sum of the $X_i$, or their average. – André Nicolas Sep 03 '15 at 17:01
  • Yes, you should be right. But just in case I misinterpreted the wording of the Question, it is as follows:

    The geometric mean of a sequence of positive numbers $x_1, . . . , x_n$ is $ (x_1 . . . x_n)^{\frac{1}{n}}$. Let$ X_1, . . . , X_n$ be a random sample on an exponential distribution with mean $θ$. We would like to use the geometric mean $G = (X_1 . . . X_n)^{\frac{1}{n}} $ to estimate $θ$. All results may be expressed in terms of the gamma function...

    Let $Y = \ln X$ where $X$ is exponential with mean $\theta$. Obtain the mean of $Y$.

    I was confused as to why there were no commas?

     – LearningMath Sep 03 '15 at 17:08
  • OK, so $X$ is exponential mean $\theta$. The relationship with $G$ is that $\ln G=\frac{1}{n}(\ln X_1+\cdots +\ln X_n)$, and since the mean of a sum is the sum of the means, $\ln G$ has mean equal to the mean of the $\ln X_i$, which is exactly what you are being asked to compute. – André Nicolas Sep 03 '15 at 17:13
  • The mean of $\ln X$ is the integral from $0$ to infinity of $(\ln x)\cdot \frac{1}{\theta}e^{-x/\theta}$. An unpleasant definite integral, which happens to have a closed form. – André Nicolas Sep 03 '15 at 17:33
  • Ok, so im a little stuck.

    I did this: $ Y = ln(X) \implies X = e^Y$ Then the CDF of $Y$, let's call this $G(y) $

    is such that $P(Y\leq y) = P(\ln(x) \leq y) = P(X \leq e^y)$

    then $G(y) = \int _0 ^{\ln(y)} \cfrac{1}{\theta} exp(-x/ \theta) dx$

    ...

    and eventually got that $E[Y] = \int _{- \infty}^{\infty} -y/\theta^2 - 1/ \theta dy$ which is infinite so I must be wrong.

    whre did I make a mistake

     – LearningMath Sep 03 '15 at 18:09
  • I had suggested a "faster" way of finding $E(y)$. But your way, the cdf of $Y$ is $\int_0^{e^y}(1/\theta)\exp(-x/\theta},dx$. We can evaluate, easily enough, then differentiate, or differentiate under the integral sign. – André Nicolas Sep 03 '15 at 18:37
  • Hi, sorry Andre, I fell asleep in the middle of last night! Your way is much more simpler, and I think what they want us to do - although I am at a standstill; We are given: $$\gamma = - \int_0 ^ \infty \ln (x) e^{-x} dx = 0.577216...$$

    using your method, we have $$E[\ln X] = \int_0 ^\infty \ln(x) \cdot \frac{1}{\theta} e^{-x/\theta}$$

    But I dont see how to massage this equation so that I can use the result of the $\gamma = \dots$ which they have given? tried a change of variables : let $x/ \theta = \tau $ then we have $\int_0 ^\infty$ $\ln (\tau \theta) 1/ \theta e^{- \tau} d(\tau \theta)$

     – LearningMath Sep 03 '15 at 23:45
  • ^ Where can I go from there? since i know have an $\ln(\tau \theta)$ instead of $\ln (\tau)$ – LearningMath Sep 03 '15 at 23:50
  • Yes, you change variables. Recall that $\ln(\tau\theta)=\ln(\tau)+\ln(\theta)$. Now it's over. – André Nicolas Sep 04 '15 at 00:38

1 Answers1

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Going through the pdf is feasible, but it is easier to note that $$E(\ln X)=\int_0^\infty \ln x\cdot\left(\frac{1}{\theta} e^{-x/\theta}\right)\,dx.$$ Let $x=u\theta$. Then $\ln(x)=\ln\theta+\ln u$ and our integral becomes $$\int_0^\infty (\ln\theta) e^{-u}\,du +\int_0^\infty (\ln x)e^{-x}\,dx.$$ The first integral is $\ln\theta$ and the second, by the (non-elementary) fact you were given is $-\gamma$.

André Nicolas
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