Gamma

I was working on some mathematical statistics problems and I came across a slick estimation problem that I thought I would share.

Problem:
Given $$\left(f\mid k\right) = \frac{x^{k-1}\exp\left(-x/\theta\right)}{\Gamma\left(k\right)\cdot\theta^{k}},\,\,\,\textrm{ where } x \gt 0,\,\,\, k \gt 0,\,\,\, \theta = 1,$$ Determine whether it is possible to find a value of $c$ such that $d\left(X\right) = cX^{2}$ will be an unbiased estimator of $k$.

My attempt at a solution:

$$E\left[c\cdot X^{2}\right] = k$$ for the left hand side to be an unbiased estimator of $k$. $$c\cdot E\left[X^{2}\right] = k$$ is true via the properties of $E\left[\cdot\right]$. Rearranging terms we can see that $$c = \frac{k}{E\left[X^{2}\right]}$$ $$c = k\cdot\left(\int_{0}^{\infty}x^{2}\frac{x^{k-1}e^{-x}}{\Gamma\left(k\right)}\mathrm{d}\,\,x\right)^{-1} = k\cdot\left(\int_{0}^{\infty}\frac{x^{k+1}e^{-x}}{\Gamma\left(k\right)}\mathrm{d}\,\,x\right)^{-1}$$ $$k\cdot\Gamma\left(k\right)\cdot\left(\int_{0}^{\infty}x^{k+1}e^{-x}\mathrm{d}\,\,x\right)^{-1} = \frac{\Gamma\left(k+1\right)}{k\cdot\Gamma\left(k+1\right)}=\frac{1}{k}.$$ So the value of $c$ such that $E\left[cX^{2}\right]$ is NOT an unbiased estimator of $k$ is $1/k$ because by definition, an unbiased estimator of some parameter $\theta$ cannot depend on the value of $\theta$.