# Sample Geometric Mean as an estimator for a Gamma Distribution

How does one compute the bias for the estimator given by the sample geometric mean for a gamma distribution with parameters ($\theta$,1)?

i.e: Given $X_1,...,X_n$ are iid with distribution Gamma($\theta$,1), what is the bias of the estimator given by: $$\hat\theta = \left(\prod_{i=1}^n X_i\right)^{1/n}$$

I have that $\mathbb{B}(\hat\theta) = \prod_{i=1}^n \mathbb{E}\left[X_i^{1/n}\right] - \theta$. Evaluating this gives an awful expression littered with gamma functions. Have I made any mistakes here?

• who cares, life is too short to do statistics, go outside, have fun, enjoy life
– Set
Oct 26, 2015 at 19:11
• so you need to find the the distribution of $X_i^{1/n}$ correct? and then evaluate the associated expected value integral, it should be fairly straight forward.
– Set
Oct 26, 2015 at 19:43

For a $\Gamma(k,\lambda)$ you have to count the integral:

$$\int_0^\infty \frac {X^{k-1+1/n}e^{-\lambda X}\lambda^k}{\Gamma(k)}dX$$

All we need is to multiply and divide the fraction by $\lambda^{1/n-1}$

$$\frac{1}{\lambda^{1/n}}\int_0^\infty \frac {(X\lambda)^{k+1/n-1}e^{-\lambda X}}{\Gamma(k)}d\lambda X$$

Which is

$$\frac {\Gamma(k+1/n)}{\lambda^{1/n}\Gamma(k)}$$

So the bias is

$$\left(\frac {\Gamma(k+1/n)}{\lambda^{1/n}\Gamma(k)}\right)^n-\theta$$

• The answer I got for $\Gamma(k,1)$ was $(\frac{\Gamma(k + \frac{1}{n})}{\Gamma(k)})^n - \theta$. Where have you got the first integral from? Oct 26, 2015 at 20:21
• It's the expectation of $X^{1/n}$. Is your answer correct? I might have made a mistake, have checked my answer one more time and didn't find anything wrong, though. Oct 26, 2015 at 20:32
• I don't have a solution unfortunately. I am confused as to how you got the term $X^{(k-1)/n}$ in the first integral though. I have used the law of the unconscious statistician and ended up with $X^{k-1+\frac{1}{n}}$ in my integral. Oct 26, 2015 at 20:49
• My god, i'm so sorry, i made a fatal mistake and will edit my answer right away. Oct 26, 2015 at 21:08
• One more time, i'm sorry. It's really late by my local time and i made a stupid mistake. Hope this helps. Oct 26, 2015 at 21:14

For a gamma random variable $X$ with shape $\alpha$ and rate $\beta$, the PDF is $$f_X (x) = \frac{\beta^\alpha x^{\alpha-1} e^{-\beta x}}{\Gamma(\alpha)}, \quad x > 0.$$ Now let $Y = g(x) = \log X$, thus $Y$ is distributed with density $$f_Y(y) = f_X(g^{-1}(y)) \left| \frac{dg^{-1}}{dy} \right| = \frac{\beta^\alpha e^{(\alpha-1)y} \exp(-\beta e^y)}{\Gamma(\alpha)} e^y = \frac{\beta^\alpha}{\Gamma(\alpha)} \exp(\alpha y - \beta e^y), \quad -\infty < y < \infty.$$ If $$\{X_1, X_2, \ldots, X_n\}$$ is a random sample from $X$, then the estimator $$\hat \theta = \left(\prod_{i=1}^n X_i\right)^{1/n} = \exp(\bar Y),$$ where $\bar Y = \frac{1}{n} \sum_{i=1}^n Y_i$. It follows that $$\operatorname{E}[\hat\theta] = \operatorname{E}[e^{\bar Y}] = M_S(1/n) = (M_Y(1/n))^n$$ where $S = n \bar Y$ and $M_S$ is the moment generating function of $S$. Now the MGF of $Y$ is trivial: $$M_Y(t) = \operatorname{E}[e^{tY}] = \int_{y=-\infty}^\infty e^{ty} f_Y(y) \, dy = \frac{\Gamma(\alpha+t)}{\Gamma(\alpha)} \frac{\beta^\alpha}{\beta^{\alpha+t}} = \frac{\Gamma(\alpha+t)}{\Gamma(\alpha)\beta^t},$$ therefore for the shape $\alpha = \theta$ and rate $\beta = 1$, we obtain $$\operatorname{E}[\hat\theta] = \left(\frac{\Gamma(\theta+1/n)}{\Gamma(\theta)}\right)^n.$$

In fact, the above is quite a bit simpler if we notice that $$M_Y(t) = \operatorname{E}[e^{tY}] = \operatorname{E}[e^{t \log X}] = \operatorname{E}[X^t] = \frac{\Gamma(\theta+t)}{\Gamma(\theta)}.$$ It is not difficult to compute the $t^{\rm th}$ moment but for some reason my brain likes to work with MGFs instead.