MHB Show that the maximum likelihood estimator is unbiased

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The maximum likelihood estimator (MLE) for the given density family is calculated as μ₁ = (1/n)Σln(Xᵢ). To establish its unbiasedness, the expectation E(μ₁) is derived, leading to E(μ₁) = (1/n)ΣE(ln(Xᵢ)). A substitution y = ln(x) is suggested to facilitate the calculation of E(ln(X)). The expectation of ln(X) is computed using an integral involving the density function, ultimately confirming that the MLE is unbiased. The discussion emphasizes the importance of recognizing the distribution of the transformed variable Y.
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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 $Y_{i}$?
 
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I'm not sure what $$f(x,\mu)$$ really is. I suppose it's

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

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

To test the unbiasness, you should calculate the expectation of $$\mu_1$$.

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

Noting $$E(\text{ln}X)=\int_0^{\infty}\text{ln}xf(x,\mu)dx=\frac{1}{\sqrt{2\pi}}\int_{\mathbb{R}}t\exp{(-(t-\mu)^2/2)}dt$$, can you figure out this?
 
stainburg said:
I'm not sure what $$f(x,\mu)$$ really is. I suppose it's

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

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

To test the unbiasness, you should calculate the expectation of $$\mu_1$$.

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

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

what substitution are you using?
 
Fermat said:
what substitution are you using?
Let $$t=\text{ln}x\in (-\infty, \infty)$$, hence $$x=\exp(t)$$.

We then have

$$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\\$$
 
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