Finding the mle of a log normal distribution

Question

So let $X1,X2,..,XN$ be an independent sample from log normal distribution with the pdf $f(x,\theta)=(x^2 \sigma^2*2\pi)^{(-1/2)}e^{-(log(x)-\theta)^2/{2\sigma^2}}$ and we have $\sigma^2=1$ and $\theta$ uknown

So I did the following we have the

$L(\theta,x)=(x_1^2\sigma^22\pi)^{-1/2}e^{-(log(x_1)-\theta)^2/{2\sigma^2}}*(x_2^2\sigma^22\pi)^{-1/2}e^{-(log(x_2)-\theta)^2/{2\sigma^2}}*...*(x_n^2\sigma^22\pi)^{-1/2}e^{-(log(x_n)-\theta)^2/{2\sigma^2}}$

and I get the following $L(\theta,x)=(2\pi)^{(-n/2)}*(\sigma^2)^{-n/2}*(1)/(x_1*x_2*..*x_n)e^{(-1/2\sigma^2)\sum(log(x_i)-\theta)^2}$

So I take the $log(L(\theta,x)$ and I get

$(-n/2)log(2\pi)+log(1/x^n)-(1/2\sigma^2)\sum(log(x_i)-\theta)^2$

So now to find the mle of $\theta$ I do $d/d(\theta)log(L(\theta,x))=0$ take the derivative

and I get $=log(x_1)-\theta+log(x_2)-\theta+...+log(x_n)-\theta$

so I get $log(x_1)+log(x_2)+..+log(x_n)-\theta*n$

so the mle $\theta[hat]$ of $\theta$ is $\theta[hat]=((log(x_1)+log(x_2)+..+log(x_n))/n$

I am not sure if this is right.

Answer 1

The question is very very simple.

But the important thing you have to keep also in mind is that the likelihood is defined unless a moltipilicative constant.

Thus if you look at your density that is

$$f(x|\theta)=\underbrace{\frac{1}{x\sqrt {2\pi}}}_{\perp \!\!\!\!\!\perp \theta}e^{-\frac{1}{2}[\log x-\theta]^2}$$

So your likelihood is the following

$$L(\theta)\propto e^{-\frac{1}{2}\Sigma_i[\log x_i-\theta]^2}$$

thus immediately you get

$$l(\theta)=-\frac{1}{2}\Sigma_i[\log x_i-\theta]^2$$

$$l^*(\theta)=\Sigma_i \log x_i -n\theta=0 \rightarrow \hat{\theta}=\frac{1}{n}\Sigma_i \log x_i$$

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concluding,

$$\hat{\theta}=\overline{Y}_n$$

where $Y_i=\log X_i$ and it is very natural knowing that, if $X\sim$ lognormal, then $\log X\sim$ normal