I have been given the following problem and starting steps but I am stuck on the first step

A measure of goodness of fit to estimate the mean $\mu$ and standard deviation $\sigma$ of a normally distributed random variable is maximum likelihood estimation. For sample $y = (y_1,... y_n)$ drawn from the normally distributed random variables, $Y_i \sim > N(\mu,\sigma)$, the likelihood is given by
$p(y|\mu,\sigma)=\prod_{i=1}^{n}\big(\frac{1}{2\pi\sigma^2}\big)^{\frac{1}{2}} exp \big( - \frac{1}{2\sigma^2}(\mu-y_i)^2 \big) $
Using the fact that $e^{-a}e^{-b} = e^{-a-b}$ for arbitrary variables a and b, show the negative log likelihood has the following form:
$L_(y|\mu,\sigma) = -\log p(y|\mu,\sigma) +\frac{1}{2\sigma^2}\sum_{i=1}^n(y_i-\mu)^2$

The steps start with splitting up the products to be
$\prod_{i=1}^n(\frac{1}{2\pi\sigma^2})^{\frac{1}{2}}\prod_{i=1}^n exp\big( - \frac{1}{2\sigma^2}(\mu-y_i)^2 \big)$

I don't see why this can be done because

$(a1 *a2) (b1* b2) \ne (a1* a2) b1 ( a1 *a2) b2$

What am I missing?


Someone explained that I could think of it this way:
Consider the sum operator
$\sum_{i=1}^n a+y_i = \sum_{i=1}^na+\sum_{i=1}^n y_i $

The product operator can work similarly
$\prod_{i=1}^n a y_i = \prod_{i=1}^na\prod_{i=1}^n y_i $

I guess I am just not used to working with the product operator.


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.