Show that the maximum likelihood estimator is unbiased

[Fermat1]

Consider a density family $f(x,{\mu})=c_{{\mu}}x^{{\mu}-1}\exp(\frac{-(\ln(x))^2)^2}{2}$ , where $c_{{\mu}}=\frac{1}{{\sqrt{2{\pi}}}}\exp(-{\mu}^2/2)$
For a sample $(X_{1},...,X_{n})$ fnd the maximum likelihood estimator and show it is unbiased. You may find the substitution $y=\ln x$ helpful.

I find the MLE to be ${\mu}_{1}=\frac{1}{n}(\ln(X_{1})+...+\ln(X_{n}))$. For unbiasedness, I'm not sure what to do. If I substitute $y_{i}=\ln(x_{i}$ I get $E({\mu}_{1})=\frac{1}{n}(E(Y_{1})+...+E(Y_{n}))$. Am I meant to recognise the distribution of the $Y_{i}$?

 

[stainburg]

I'm not sure what \(\displaystyle f(x,\mu)\) really is. I suppose it's

\(\displaystyle f(x,\mu)=c_{\mu}x^{\mu-1}\exp(-(\text{ln}x)^2/2)\).

Give a sample \(\displaystyle (X_1,X_2...,X_n)\), and you've got the MLE, \(\displaystyle \mu_1=\frac{1}{n}\sum_{i=1}^{n}\text{ln}X_i\). For this \(\displaystyle f(x,\mu)\), that's right.

To test the unbiasness, you should calculate the expectation of \(\displaystyle \mu_1\).

Thus, we have, \(\displaystyle E(\mu_1)=\frac{1}{n}\sum_{i=1}^nE(\text{ln}X_i)\).

Noting \(\displaystyle E(\text{ln}X)=\int_0^{\infty}\text{ln}xf(x,\mu)dx=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}t\exp{(-(t-\mu)^2/2)}dt\), can you figure out this?

 

[Fermat1]

I'm not sure what \(\displaystyle f(x,\mu)\) really is. I suppose it's

\(\displaystyle f(x,\mu)=c_{\mu}x^{\mu-1}\exp(-(\text{ln}x)^2/2)\).

Give a sample \(\displaystyle (X_1,X_2...,X_n)\), and you've got the MLE, \(\displaystyle \mu_1=\frac{1}{n}\sum_{i=1}^{n}\text{ln}X_i\). For this \(\displaystyle f(x,\mu)\), that's right.

To test the unbiasness, you should calculate the expectation of \(\displaystyle \mu_1\).

Thus, we have, \(\displaystyle E(\mu_1)=\frac{1}{n}\sum_{i=1}^nE(\text{ln}X_i)\).

Noting \(\displaystyle E(\text{ln}X)=\int_0^{\infty}\text{ln}xf(x,\mu)dx=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}t\exp{(-(t-\mu)^2/2)}dt\), can you figure out this?

what substitution are you using?

 

[stainburg]

what substitution are you using?

Let \(\displaystyle t=\text{ln}x\in (-\infty, \infty)\), hence \(\displaystyle x=\exp(t)\).

We then have

\(\displaystyle E(\text{ln}X)\\

=\int_0^{\infty}\text{ln}xc_{\mu}x^{\mu-1}\exp(-(\text{ln}x)^2/2)dx\\

=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}\exp(-\mu^2/2)t\exp{((\mu-1)t)}\exp{(-t^2/2)}d(\exp(t))\\

=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}t\exp{(-(t^2-2\mu t+\mu^2)/2)}dt\\

=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}t\exp{(-(t-\mu)^2/2)}dt\\\)

 

[blue_raver22]

Yes, recognizing the distribution of the $Y_{i}$ is an important step in showing the unbiasedness of the MLE. In this case, the $Y_{i}$ are normally distributed with mean $\mu$ and variance 1. Using the properties of the normal distribution, we can see that $E(Y_{i})=\mu$ and $Var(Y_{i})=1$. Therefore, $E({\mu}_{1})=\frac{1}{n}(n\mu)=\mu$, which shows that the MLE is unbiased. This means that on average, the MLE will give us the true value of $\mu$ as the estimate.