# How can you have a density function that depends on a parameter?

Let $X_1,...,X_n$ be a random sample from a distribution with mass/density function $f_X$ that depends on a (possible vector) parameter $\theta$. Then $f_{X_1}(x_1) = f_X(x_1;\theta)$ so that

$f_{X_1,...,X_k}(x_1,...,x_k) = \prod_{i=1}^kf_X(x_i;\theta).$

Could someone please explain what the significance of $\theta$ is in the above definition. I've never seen this before. Is it the mass/density function that depends on the parameter or is it the random sample that depends on the parameter.

-

## 1 Answer

θ is called a parameter. It is an unknown constant that is a part of the formula for the density function. So for example if you have a normal distribution with variance 1 and unknown mean the density would be fx(x1;θ) = [exp(-(x1-θ)$^2$ /2)/√(2π)]. θ is the unknown mean in this case. This is very common in statistics. You will need to get familiar with things like this if you want to be able to follow what is being said on this site about statistics.

-
And after you get familiar with that, then you can try "non-parametric statistics"... –  GEdgar May 20 '12 at 21:26
@user1066113 Nonparametric statistics does not have parametric families like this. So actually you could go to nonparametrics with out getting used to these parametric families. –  Michael Chernick May 20 '12 at 21:32
This is a good way of thinking, but it seems a tad Statistics-centric and very informal mathematically, though I'll grant that it is difficult to explain this formally at the level of an average individual who asks this question. –  guy May 20 '12 at 21:32
@guy Since I am a statistician (PhD in statistics MA in mathematics) that may explain the statistical orientation of my answer. But the OP asked what the significance of the parameter theta was in the likelihood equation. It conveyed the impression that the OP needed a definition for a parametric density. Perhaps I should have added that by differentiating the likelihood with respect to theta one can solve for the value of theta that maximimizes the function. That would be a more complete answer and mixes a little math with a little statistics. –  Michael Chernick May 20 '12 at 22:27