MLE for $\mu$ for Normal distribution

Question

I am doing some homework and have a little trouble with the following problem:

Suppose the weight $X$ in pounds of a girl in a certain town is normally distributed with $N(\mu,15^2)$ while the weight $Y$ of a boy in this town is normally distributed $N(1.3 \mu, 20^2).$ The weights of randomly chosen girl and boy are $x = 95$ and $y = 130$ pounds respectively.

Then I have to find the MLE of $\mu$.

However, I know that the MLE of $\mu$ for a normal distribution is simply the sample mean, thus

$$MLE_{\mu} = \frac{1}{n} \sum_{i = 1}^n X_i = \frac{95 + 130}{2} = 112.5$$

Is this simply it?

TIA for any help.

Answer 1

It is not this simple.

You have got two singleton samples from two different normal populations $N(\mu,15^2),N(1.3\mu,20^2)$. The joint distribution of the weight of a boy and girl is

$$l(x,y)=\frac{\exp\left(-\frac{(x-\mu)^2}{2(15^2)}-\frac{(y-1.3\mu)^2}{2(20^2)}\right)}{2\pi(15)(20)}$$

This likelihood function attains maximum at $$\hat\mu_{MLE}=\frac{\frac x{15^2}+\frac{1.3y}{20^2}}{\frac1{15^2}+\frac{1.3^2}{20^2}}$$

Answer 2

The trick to find $\hat{\mu}$ is to expand the exponent, waste any quantity not depending on $\mu$ and factorize the exponent in order to have again a perfect square (adding and subtracting a customized quantity).

$$L\propto \exp\left\{-\frac{1}{2}\left[ \frac{(x-\mu)^2}{15^2} + \frac{(y-1.3\mu)^2}{20^2}\right]\right\}=$$

$$=\exp\left\{ -\frac{1}{2}\left[ \mu^2\left(\frac{1}{15^2}+\frac{1.3^2}{20^2} \right) -2\mu \left( \frac{x}{15^2}+\frac{1.3y}{20^2} \right)\right] \right\}=$$

$$=\exp\left\{ -\frac{1}{2\frac{1}{\frac{1}{15^2}+\frac{1.3^2}{20^2}}}\left[ \mu^2 -2\mu \frac{ \frac{x}{15^2}+\frac{1.3y}{20^2}} {\frac{1}{15^2}+\frac{1.3^2}{20^2}}\pm ?\right] \right\}$$

now if you look at the quantity in the square brackets, you will recognize an incomplete square expression. After completing it, in the brackets you will have an expression like

$$[\mu-\theta]^2$$

and thus the MLE is the quantity near $2\mu$, corresponding to the quantity written by @Shubham Johri (+1)