Not able to prove unbiasedness of estimator

Question

I am taking $N$ i.i.d. samples ${y}_{1},{y}_{2},\dots,{y}_{N}$ from the following distribution:

$f\left( y,\theta \right) =\begin{cases} \frac { { y }^{ 2 } }{ 2{ \theta }^{ 3 } } { e }^{ -\frac { y }{ \theta } }, & y\ge 0 \\ 0, & y<0 \end{cases}$

where $\theta > 0$.

I have computed its Maximum Likelihood estimator: ${ \widehat { \theta } }_{ ML }=\frac { 1 }{ 3N } \sum _{ i=1 }^{ N }{ { y }_{ i } }$

I am trying to prove that ${ \widehat { \theta } }_{ ML }$ is unbiased, however, I get to the following:

$E\left[ { \widehat { \theta } }_{ ML } \right] =E\left[ \frac { 1 }{ 3N } \sum _{ i=1 }^{ N }{ { y }_{ i } } \right] =\frac { 1 }{ 3N } E\left[ \sum _{ i=1 }^{ N }{ { y }_{ i } } \right] =\frac { 1 }{ 3N } \sum _{ i=1 }^{ N }{ E\left[ { y }_{ i } \right] } =\frac { 1 }{ 3N } \sum _{ i=1 }^{ N }{ \mu } =\frac { N }{ 3N } \mu =\frac { 1 }{ 3 } \mu $

How could this be further developed as to prove its unbiasedness?

I am trying to write $\mu$ in terms of $\theta$, but I cannot figure out how...

Answer 1

You have proved that $E[\hat \theta_{MLE}] = \dfrac{\mu}{3}$. But you want to prove $E[\hat \theta_{MLE}] = \theta$.

The question remains: Is $\dfrac{\mu}{3} = \theta$? Or equivalent $\mu = E[Y] = 3\theta$?

Calculating $E[Y]$ through $E[Y] = \int_{-\infty}^{+\infty}yf_Y(y; \theta) \operatorname d y$ yields:

$$ \begin{align} E[Y] = \int_{-\infty}^{+\infty}yf_Y(y; \theta) \operatorname d y & = \int_0^{+\infty}\dfrac{y^3}{2\theta^3}e^{\frac{-y}{\theta}} \operatorname d y \\ & = \frac{\theta}{2}\int_{0}^{+\infty}t^3e^{-t} \operatorname d t\\ & = \frac{\theta}{2} \Gamma(4) \\ & = \frac{\theta}{2} \cdot 3! = 3\theta \end{align} $$

using variable substitution and the gamma function $(n-1)! = \Gamma(n) = \int_0^{+\infty} x^{n-1} e^{-x}\operatorname d x$.

edit

If you haven't learned about the gamma function, partial integration can be used to evaluate the integral. The gamma function is handy shortcut.