MLE for $\mu$ for Normal distribution 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.
 A: 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 $$\hat\mu_{MLE}=\frac{\frac x{15^2}+\frac{1.3y}{20^2}}{\frac1{15^2}+\frac{1.3^2}{20^2}}$$
A: 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)
