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, $$Var[g(x)] = E\left[g(x)^2\right] - \left(E[g(x)]\right)^2 = \int{g(x)^2 f(x) dx} - \left(\int {g(x)f(x)dx}\right)^2$$

where each integral is over the domain of x.

 

[blue_raver22]

To find the expected value (E) of ln(x), we can use the definition of expected value which is the integral of the function multiplied by the probability density function. In this case, we have:

E(ln x) = \int_{0}^{1} ln(x) \theta x^{\theta - 1} dx

To solve this integral, we can use integration by parts. Let u = ln(x) and dv = \theta x^{\theta - 1} dx. Then, du = \frac{1}{x} dx and v = x^{\theta}. Substituting these into the integral, we get:

E(ln x) = \left[ln(x) x^{\theta}\right]_{0}^{1} - \int_{0}^{1} \frac{x^{\theta}}{x} dx

E(ln x) = \left[ln(x) x^{\theta}\right]_{0}^{1} - \int_{0}^{1} x^{\theta - 1} dx

E(ln x) = \left[ln(x) x^{\theta}\right]_{0}^{1} - \left[\frac{x^{\theta}}{\theta}\right]_{0}^{1}

E(ln x) = 1 - \frac{1}{\theta}

To find the variance (Var) of ln(x), we can use the definition of variance which is the expected value of the squared difference between the random variable and its mean. In this case, we have:

Var(ln x) = E[(ln x - E(ln x))^2]

Substituting the value we found for E(ln x), we get:

Var(ln x) = E[(ln x - 1 + \frac{1}{\theta})^2]

Using the same method as above, we can solve this integral to get:

Var(ln x) = \frac{2}{\theta^2}

Now, to apply the Cramer-Rao lower bound, we need to find the Fisher information (I) for the given distribution. The Fisher information is defined as:

I = -E\left[\frac{\partial^2}{\partial \theta^2} ln f(x; \theta)\right]

Where f(x; \theta) is the probability density function. In this case, we have:

f(x; \theta) =