Finding E(ln x) and Var(ln x): Cramer-Rao Lower Bound

[safina]

May I ask how to find the E\left(ln x\right) and Var\left(ln x)?
The X_{i} are random sample from the f\left(x; \theta\right) = \theta x^{\theta - 1}I_{\left(0, 1\right)}\left(x\right) where \theta > 0.

I need the information in finally solving the Cramer-Rao lower bound for the variance of an unbiased estimator of a function of \theta. And also for checking if I have an unbiased estimator.

 

[SW VandeCarr]

May I ask how to find the E\left(ln x\right) and Var\left(ln x)?

A random variable X has a lognormal distribution if X=e^{Y} where Y is normally distributed with mean \mu and standard deviation \sigma.

Then: E(X)=e^{\mu+\sigma/2}, Var(X)=e^{2(\mu+\sigma)}-e^{2\mu+\sigma}

The moment generating function is: E(X^n)=e^{n\mu+(n^2\sigma^2/2)}

Note the moments of the lognormal distribution do not uniquely define the distribution and a finite mgf only exists on the interval (-\infty,0]

EDIT:: You can assign other distributions to Y, but without specification, I don't think there's a single answer to your question. The lognormal is the default under the Central Limit Theorem..

 

[EnumaElish]

Let g(x)= ln x. For E[g(x)] see the inner product definition here. For Var[g(x)], look under continuous case here with g(x) replacing x and \mu is interpreted as E[g(x)]; alternatively see the delta method.

 

[EnumaElish]

That is, Var[g(x)] = integral of ( g(x) - E[g(x)] )^2 f(x), over x's domain.

Alternatively, <br /> <br /> Var[g(x)] = E\left[g(x)^2\right] - \left(E[g(x)]\right)^2 <br /> = \int{g(x)^2 f(x) dx} - \left(\int {g(x)f(x)dx}\right)^2<br /> <br />

where each integral is over the domain of x.