I have -before- found the MLE of the two parameters of a Normal Distribution but I don't have any idea about how to proceed in this case.


A sample of size $n$ is drawn from each of four normal populations, all of which have the same variance $\sigma^2$. The means of the four populations are $a + b + c$, $a + b - c$, $a - b + c$, and $a - b - c$.

What are the maximum-likelihood estimators of $a, b, c$, and $\sigma^2$?

(The sample observations may be denoted by $X_{ij}$, $i = 1, 2, 3,4$ and $j = 1,2, ... , n$.)


2 Answers 2


Hint: Notice that the distribution is perfectly symmetric around the value $a$. That means the maximum-likelihood estimator of $a$ will be simply the mean of the data.

Can you continue?

enter image description here


Here, the likelihood function is $$L(\theta; X_{ij}) = L(\theta; X_{11}, X_{12}, … , X_{1n}, ~ X_{21}, X_{22}, … , X_{2n}, ~X_{31},X_{32}, …, X_{3n}, ~X_{41},, X_{42}, … , X_{4n})$$ where $\theta$ is the variable that you want to minimize w.r.t.

Since the $X_{ij}$'s are iid, and since minimizing the log of the function will yield the same result, we want to minimize $$\ln L(\theta; X_{ij}) = \ln \prod_{ij} p(X_k;\theta)$$ where $p(X_i;\theta)$ is the distribution of each of the $X_{ij}$'s. So we have

\begin{align*} \ln L(\theta; X_{ij}) & = \ln \prod_{ij} p(X_k;\theta) \\ & = \sum_{ij} \ln{p(X_k;\theta)} \\ & = \sum_{i=1}^n \ln{\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(X_{1i}-(a+b+c))^2}{2\sigma^2}}} +\sum_{i=1}^n \ln{\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(X_2i-(a+b-c))^2}{2\sigma^2}}} +\sum_{i=1}^n \ln{\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(X_3i-(a-b+c))^2}{2\sigma^2}}} +\sum_{i=1}^n \ln{\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(X_4i-(a-b-c))^2}{2\sigma^2}}} \\ & = \sum_{i=1}^n \left[ \ln{\frac{1}{\sqrt{2\pi\sigma^2}} + \ln e^{-\frac{(X_{1i}-(a+b+c))^2}{2\sigma^2}}}\right] + \text{similarly for the other 3 sums} \\ & = n \ln{\frac{1}{\sqrt{2\pi\sigma^2}}} - \sum_{i-1}^n \frac{[X_{1i}-(a+b+c)]^2}{2\sigma^2} + \text{similarly for the other 3 sums} \\ \end{align*}

Lastly you want to minimize this (by taking the derivative and setting to equal to zero) w.r.t. each of the variables at a time to get their respective MLE's. For instance,

\begin{align*} \frac{\mathrm{d} }{\mathrm{d} a}\ln L(a; X_{ij}) & = \sum_{i=1}^n \frac{[X_{1i}-(a+b+c)]}{\sigma^2} +\sum_{i=1}^n \frac{[X_{2i}-(a+b-c)]}{\sigma ^2} +\sum_{i=1}^n \frac{[X_{3i}-(a-b+c)]}{\sigma ^2} + \sum_{i=1}^n \frac{[X_{4i}-(a-b-c)]}{\sigma ^2}\\ & = \frac{n}{\sigma ^2}(\bar{X_1}-(a+b+c)) + \frac{n}{\sigma ^2}(\bar{X_2}-(a+b-c)) +\frac{n}{\sigma ^2}(\bar{X_3}-(a-b+c)) + \frac{n}{\sigma ^2}(\bar{X_4}-(a-b-c)) \\ & = \frac{n}{\sigma ^2} \left[\bar{X_1}+\bar{X_2}+\bar{X_3}+\bar{X_4} -4a \right] \end{align*}

So the MLE of a is when this is equal to zero: $\frac{n}{\sigma ^2} \left[\bar{X_1}+\bar{X_2}+\bar{X_3}+\bar{X_4} -4a \right] = 0 \implies a = \frac {\bar{X_1}+\bar{X_2}+\bar{X_3}+\bar{X_4}}{4}.$

  • $\begingroup$ Just one question, how do you knnow they are iid? $\endgroup$ Jun 2, 2019 at 21:21
  • $\begingroup$ Why did you use the word minimize? Don't we have to maximize? $\endgroup$
    – Daman
    Jan 17, 2021 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.