How to use method of moments to estimate mean?

Question

I have given the density function $$f(\epsilon,\mu) = (1-\epsilon)\frac{1}{3}e^{-x/3}+\epsilon\frac{1}{\mu}e^{-x/\mu}.$$ And would like to estimate $\mu$ using the method of moments and then numerically maximize the likelihood as a function of $\epsilon$. I know that mgf is $$M_n(x) = E[(X-EX)^n],$$ but I don't know how to calculate this to use method of moemnts in thus task. Will appreciate a hint, thank you!

Answer 1

Since $f$'s support being $[0,\,\infty)$ implies$$E[X^n]=n!((1-\epsilon)3^n+\epsilon\mu^n),$$we get $(1-\epsilon)3+\epsilon\mu=M_1$ and $(1-\epsilon)9+\epsilon\mu^2=M_1^2+M_2$. In particular,$$\epsilon\mu(3-\mu)=3M_1-M_1^2-M_2$$is a quadratic in $\mu$ we can solve to express it in terms of $M_1,\,M_2,\,\epsilon$ (just make sure you take $\mu>0$). MLE is then equivalent to maximizing the sample mean of $\ln f$ as a function of $\epsilon$, after we've removed its explicit $\mu$-dependence by substitution.

Answer 2

First, I'm rusty. Second, I haven't seen all the notation. But I'm a little suspicious of your bit above on moment-generating functions, my primary concern being that the RHS looks a lot like the $n$th central moment.

Anyway.

If $a_1, a_2, \ldots a_n$ are a sequence of non-negative numbers summing to $1$ and $f_1, f_2, \ldots f_n$ are pdfs, then we can define a new pdf by $ f(x) = \sum_{i \le n} a_i f_i (x) $. The linearity of the definite integral means that we'll also have, for example, $EX^n = \sum_{i\le n} a_i E_i[X^n]$, where each $E_i$ is the expectation taken with respect to $f_i$.

Assuming the support for your distribution is non-negative values of $x$, your pdf is this type of mixture, specifically one comprising two exponential distributions, one with mean $3$ and another with unknown mean $\mu$. If $W \sim \exp(\lambda)$, then $EW = \lambda$ and $EW^2 = 2\lambda^2 $. Therefore, we have

\begin{aligned} \ \\ EX &= (1 - \epsilon)\cdot 3 + \epsilon\cdot\mu \ \\ EX^2 &= (1 - \epsilon)\cdot 18 + \epsilon\cdot 2\mu^2. \end{aligned}

Using the method of moments, we'd set the above expressions equal to $\bar{x}$ and $\frac{1}{n}\sum x_i^2$. So then

\begin{aligned} \ \\ \bar{x} &= 3(1 - \epsilon) + \epsilon\mu \ \\ \frac{1}{n}\sum x_i^2 &= 18(1 - \epsilon) + 2\epsilon\mu^2. \end{aligned}

From here, we can solve for both $\mu$ and $\epsilon$ in terms of the known values of $\bar{x}$ and $\frac{1}{n}\sum x_i^2$.