Is $\hat{\mu}$ and unbaiased estimator of $\mu?$

Question

Problem: Let $X_1,...,X_n$ be a sample from a distribution with CDF

$$F(x)=1-\frac{1}{(1+x)^\mu}, \quad x>0, \quad \mu > 0.$$

Find the maximum likelihood $\hat{\mu}$ of $\mu.$ Also, determine if this estimator is unbaiased.

I have that

$$\hat{\mu}=\frac{n}{\sum_{k=1}^n\ln(1+X_k)}.$$

Which is correct from the given problem. In order for $\hat{\mu}$ to be an unbaiased estimator then it has to be the case that $E[\hat{\mu}]=\mu.$ The solution says we can do the following:

It is not an unbaiased estimator. To show this, we can let $n=\mu=1.$ We then have

$$E[\hat{\mu}]\int_0^{\infty}P(X>x)\ dx=\int_0^{\infty}\frac{1}{1+x} \ dx=\infty\neq\mu.$$

But I thought that

$$E[\hat{\mu}]=E\left[\frac{1}{\sum_{k=1}^1\ln(1+X_1)}\right]=E\left[\frac{1}{\ln{(1+X_1)}}\right]=\int_0^{\infty}\frac{1}{\ln{(1+x)}}\cdot\frac{1}{x+1} \ dx, $$

in accordance with LOTUS? Where does he get $P(X>x)$ all of a sudden?

score 1 · Accepted Answer · answered May 30 '18 at 22:24

1

It is a well known fact that for any non-negative random variable $X$,

$$ E[X] = \int_0^\infty P(X > x)dx $$

What is the distribution of $X$ here though?

answered May 30 '18 at 22:24

Thanks. I edited the question to include the entire problem statement. I still don't see how $P(X>x)=\frac{1}{x+1}?$ – Parseval May 30 '18 at 22:34
That does seem like a mistake. I'm assuming they meant to show that $X$ itself is not an unbiased estimator of $\mu$. Your expression for $E[\hat{\mu}]$ is correct. – May 30 '18 at 23:02

score 1 · Answer 2 · answered May 31 '18 at 03:47

Answer by Flowsnake correctly points to expected value of a non-negative random variable being summing/integrating the survival function $1-F(x)$ which in this case is integrating $1-\left[1-\frac{1}{(1+x)}\right]=\frac{1}{1+x}$. That is, the mean $\mu = EX$ here is

$$\mu = \int_0^{\infty} \left[1-F(x)\right] dx = \int_0^{\infty} \frac{1}{1+x} dx $$

which is a divergent integral.

Regarding your question about applying the LOTUS (Law of the Unconscious Statistician): It requires finding $E(g(X))$ by integrating $g(x)$ times the PDF $f(x)$ which would mean $E\left[\frac{1}{\ln{(1+X_1)}}\right]=\int_0^{\infty}\frac{1}{\ln{(1+x)}}\cdot \mathbf{f(x)} \, dx$. But the PDF $\mathbf{f(x)}$ here would be

$$f(x) = F'(x) = \left[1-\frac{1}{1+x}\right]' = \frac{1}{(1+x)^2}$$

so that

\begin{align} E[\hat{\mu}] &=E\left[\frac{1}{\ln{(1+X_1)}}\right] \\ &=\int_0^{\infty}\frac{1}{\ln{(1+x)}}\cdot\frac{1}{(1+x)^2} \ dx \end{align}

Is $\hat{\mu}$ and unbaiased estimator of $\mu?$

2 Answers2