# Mean and Variance problem

• Jul 18th 2010, 02:26 PM
firebio
Mean and Variance problem
Let T= $\sum ln(1+x_i)$

Find the mean and variance of T.

some other info: $X_1,X_2....X_n$ are iid rv with density function $f(x,h)= h(1+x) ^\ (-h-1)$, x>=0 and h>0, h is an unknown variable

I'm not really sure how to approach this problem
Any help will be appreciated

• Jul 18th 2010, 03:33 PM
SpringFan25
Obtain the mean and variance of any particular ln(1+x):

Define $Y_i=ln(1+x_i)$
$E(Y_i) = \int f(x_i,h)ln(1+x) dx_i$

$E^2(Y_i) = \int f(x,h)\left( ln(1+x)\right) ^2 dx$

$Var(Y_i) = E^2(Y_i) - \left( E(Y_i) \right)^2$

I haven't checked but those integrals look do-able to me.

Then,Use the standard results for summing independant variables
$E \left( \sum \left(Y_i \right) \right)= \sum E \left(Y_i \right)$...

$Var \left( \sum \left(Y_i \right) \right)= \sum Var \left(Y_i \right)$
• Jul 18th 2010, 05:59 PM
firebio
I got

$E(Y_i) = {-(x+1)^ {-h} (h(ln(x+1)+1) * \frac {1}{h}$
$E(Y_i)^2 = {-(x+1) ^ {-h} (h^2 (ln^2(x+1) +2h ln (x+1)+2)) * \frac {1}{h^2}$
from wolfram

For $\sum E(Y_i)$, it is just ${-\sum(x+1)^ {-h} (h \sum(ln(x+1) +1) * \frac {1}{h}$
• Jul 20th 2010, 05:04 AM
SpringFan25
are you supposed to use wolfram to do your integrals?

anyway, for the last step you are expected to notice that all the $Y_i$ are iid, so

$\displaystyle \sum_{i=1}^{n} E(Y_i) = nE(Y)$

$\displaystyle \sum_{i=1}^{n} Var(Y_i) = nVar(Y)$