# How to find unbiased estimator when a sample is drawn from a pdf which is a linear combination of pdfs of two normal distributions?

Let $$X_1,X_2,\ldots,X_n$$ be a random sample from a continuous distribution with the probability density function $$f(x)=\frac{1}{2\sqrt{2\pi}}\left[e^{-\frac12(x-2\mu)^2}+e^{-\frac12(x-4\mu)^2}\right],\quad-\infty

If $$T=\sum\limits_{i=1}^n X_i$$, then which one of the following is an unbiased estimator of $$\mu$$?

$$(A)\quad\frac{T}{n}\qquad\quad(B)\quad\frac{T}{2n}\qquad\quad(C)\quad\frac{T}{3n}\qquad\quad(D)\quad\frac{T}{4n}$$

If a function has a PDF that is the linear combination of two normal distribution, how to find its unbiased estimator?

I know the unbiased estimator of normal distribution is $$\mu$$ but in linear combination I don't know.

• Please use MathJax for typesetting math here. – StubbornAtom Oct 2 '19 at 14:53

The key is to recognize that if $$Z_i$$ is a Bernoulli($$1/2$$) random variable, and the conditional distribution of $$X_i$$ given $$Z_i=z$$ is $$N(2\mu, 1)$$ if $$z=0$$, and $$N(4\mu, 1)$$ if $$z=1$$, then the marginal PDF of $$X_i$$ is the given density $$f$$.
Using the tower property, we have $$\mathbb{E}[X_i] = \mathbb{E}[\mathbb{E}[X_i \mid Z_i]] = \frac{1}{2} (2 \mu) + \frac{1}{2} (4\mu) = 3 \mu$$. From here you can compute $$\mathbb{E}[T]$$.