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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 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...

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: ${ \widehat { \theta } }_{ ML }=-\frac { 1 }{ 3 } \sum _{ i=1 }^{ N }{ { y }_{ i } }$

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

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

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

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?