Let $X_{1},...,X_{n}$ be independent random variables with $X_{k}$ having the normal distribution with mean $k\theta$ and variance $\sigma^2$. Find the maximum likelihood estimator of $\theta$.

My approach

The likelihood function of the given density can be written as:

$L(X|\theta,\sigma^2)=\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2\sigma^2}(x_{i}-\theta)^2}\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2\sigma^2}(x_{i}-2\theta)^2}....\frac{1}{\sigma \sqrt{2\pi}}e^{\frac{-1}{2\sigma^2}(x_{i}-n\theta)^2}$

After solving for $\theta$, the maximum likelihood estimator of $\theta$ is obtained as follows:


I have not solved this kind of problem in maximum likelihood estimation and the likelihood equation was not in very easy form. Also, the manual didn't give me the answer for the problem. I want to make sure whether my steps were correct or not.

Thanks in advance.


Your steps are right, but you have an algebra error. The negative log-likelihood for $\theta$ is \begin{align*} -\log L(\theta | X) &= \dfrac{1}{2\sigma^2} \sum_{i=1}^{n}(X_i - i\theta)^2\\ \dfrac{-\log L(\theta | X)}{d \theta} & = -\dfrac{1}{\sigma^2} \sum_{i=1}^{n}i(X_i - i\theta) \overset{set}{=} 0\\ \Rightarrow \hat{\theta} &= \dfrac{6 \sum_{i=1}^{n} i X_i}{n (n+1)(2n+1)} \end{align*}

You may have missed cancelling the 2 when taking the derivative. A good practice is to check the answer using a simulation, since the MLE should be consistent in this case.

> # Checking answers
> set.seed(100)
> n <- 100
> theta <- 5
> sig <- 2
> x <- rnorm(n, (1:n)*theta, sig)
> (theta.hat <- 6*sum((1:n)*x)/(n*(n+1)*(2*n+1)))
[1] 4.998773
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  • $\begingroup$ thank you so much for reviewing my steps. I will be careful while taking the log derivative next time. $\endgroup$ – user8125394 Feb 2 '18 at 10:02

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