maximum likelihoodnormal distributionprobability distributions
I have the pdf of a log-normal distribution:
$$ f(y;\theta)= \frac {1}{y\sqrt{2\pi\theta}}\exp \left(-{\frac {(\log y)^2}{ 2\theta}}\right)$$
for $y>0$ and $\theta>0$ and $f(y; \theta) = 0$ otherwise.
and assuming that $E(\log Y_i) = 0$ and $\operatorname{var}(\log Y_i) = \theta$
I need help finding the maximum likelihood estimator of $\theta$ so I can find the expected information and the approx standard error.
Can I please get some help on how to get started?
The likelihood function is $$L(\theta|\mathbb x)=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_i \le 2\theta ,\forall i\\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$ $$=\begin{cases}\dfrac{1}{\theta ^n},\,\,\,\theta \le x_{(1)} \le x_{(n)} \le2\theta \\0,\,\,\,\,\,\,\,\,\text{otherwise}\end{cases}$$ For $\theta \ge \dfrac {x_{(n)}}{2},L(\theta|\mathbb x)=\dfrac{1}{\theta ^n}$ is a decreasing function in $\theta$. Hence MLE of $\theta$ is $\color{blue}{\hat\theta=\dfrac{X_{(n)}}{2}}$.
Additional comments: Your answer seems OK. It may be of interest to know that $\hat \theta$ is not unbiased. One can get a rough idea of the distribution of $\hat \theta$ for a particular $\theta$ by simulating many samples of size $n.$ I don't know of a convenient 'unbiasing' constant multiple. The Wikipedia article I linked in my Comment above gives more information.
Here is a simulation for $n = 10$ and $\theta = 5.$
th = 5; n = 10
th.mle = -n/replicate(10^6, sum(log(rbeta(n, th, 1))))
mean(th.mle)
## 5.555069 # aprx expectation of th.mle > th = 5.
median(th.mle)
## 5.172145
The histogram below shows the simulated distribution of $\hat \theta.$ The vertical red line is at the mean of that distribution, and the green curve is its kernel density estimator (KDE). According to the KDE, its mode is near $4.62.$
den.inf = density(th.mle)
den.inf$x[den.inf$y==max(den.inf$y)]
## 4.624876
hist(th.mle, br=50, prob=T, col="skyblue2", main="")
abline(v = mean(th.mle), col="red")
lines(density(th.mle), lwd=2, col="darkgreen")
Addendum on Parametric Bootstrap Confidence Interval for $\theta:$
In order to find a confidence interval (CI) for $\theta$ based on MLE $\hat \theta,$ we would like to know the distribution of $V = \frac{\hat \theta}{\theta}.$ When that distribution is not readily available, we can use a parametric bootstrap.
If we knew the distribution of $V,$ then we could find numbers $L$ and $U$ such that $P(L \le V = \hat\theta/\theta \le U) = 0.95$ so that a 95% CI would be of the form $\left(\frac{\hat \theta}{U},\, \frac{\hat\theta}{L}\right).$ Because we do not know the distribution of $V$ we use a bootstrap procedure to get serviceable approximations $L^*$ and $U^*$ of $L$ and $U.$ respectively.
To begin, suppose we have a random sample of size $n = 50$ from $\mathsf{Beta}(\theta, 1)$ where $\theta$ is unknown and its observed MLE is $\hat \theta = 6.511.$
Entering, the so-called 'bootstrap world'. we take repeated 're-samples` of size $n=50$ from $\mathsf{Beta}(\hat \theta =6.511, 0),$ Then we we find the bootstrap estimate $\hat \theta^*$ from each re-sample. Temporarily using the observed MLE $\hat \theta = 6.511$ as a proxy for the unknown $\theta,$ we find a large number $B$ of re-sampled values $V^* = \hat\theta^2/\hat \theta.$ Then we use quantiles .02 and .97 of these $V^*$'s as $L^*$ and $U^*,$ respectively.
Returning to the 'real world' the observed MLE $\hat \theta$ returns to its original role as an estimator, and the 95% parametric bootstrap CI is $\left(\frac{\hat\theta}{U^*},\, \frac{\hat\theta}{L^*}\right).$
The R code, in which re-sampled quantities are denoted by .re instead of $*$, is shown below.
For this run with set.seed(213) the 95% CI is $(4.94, 8.69).$ Other runs with unspecified
seeds using $B=10,000$ re-samples of size $n = 50$ will give very similar values. [In a real-life application, we would not know whether this CI covers the 'true' value of $\theta.$ However,
I generated the original 50 observations using parameter value $\theta = 6.5,$ so in this demonstration we
do know that the CI covers the true parameter value $\theta.$ We could have used the
probability-symmetric CI with quantiles .025 and .975, but the one shown is a little shorter.]
set.seed(213)
B = 10000; n = 50; th.mle.obs=6.511
v.re = th.mle.obs/replicate(B, -n/sum(log(rbeta(n,th.mle.obs,1))))
L.re = quantile(v.re, .02); U.re = quantile(v.re, .97)
c(th.mle.obs/U.re, th.mle.obs/L.re)
## 98% 3%
## 4.936096 8.691692
Best Answer
So the log likelihood given $n$ observations is $$l(\theta) = \log \prod_{i=1}^n f(y_i;\theta) \propto - \sum_{i=1}^n \log y_i - \frac{n}{2} \log \theta - \frac{1}{2\theta}\sum_{i=1}^n \log^2 y_i$$ Derive w.r.t $\theta$ and set it to zero to get the first order condition: $$l'(\theta) = - \frac{n}{2\hat{\theta}} + \frac{1}{2\hat{\theta}^2} \sum_{i=1}^n \log^2 y_i = 0$$ or $$ n = \frac{1}{\hat{\theta}} \sum_{i=1}^n \log^2 y_i$$ This means that the MLE of $\theta$ is $$ \hat{\theta} = \frac{1}{n} \sum_{i=1}^n \log^2 y_i$$