I am making some kind of systematic error(s), while working with maximum likelihood estimations. Could someone please point these out to me?
In my last assignment, I tried to find the MLE of $\beta$ for a gamma distributed variable.
As follows:
The density function is
$$f(x)=\frac{1}{\theta^2}ye^{(\frac{-y}{\theta})}$$
This can be separated into:
$$f(x)=\frac{1}{\theta^2}*\frac{y}{1}*e^{(\frac{-y}{\theta})}$$
The probability for th $n^{th}$ outcome can be written as
$$p(y)=(\frac{1}{\theta^2}*\frac{y_1}{1}*e^{(\frac{-y_1}{\theta})})*(\frac{1}{\theta^2}*\frac{y_2}{1}*e^{(\frac{-y_2}{\theta})})....(\frac{1}{\theta^2}*\frac{y_n}{1}*e^{(\frac{-y_n}{\theta})})$$
$$p(y)=(\frac{1}{\theta^2})^n*\prod (\frac{y_i}{1}*e^{(\frac{-y_i}{\theta})})$$
We convert this expression to logarithms:
$$p(y)=ln (\frac{n}{\theta^2}) + ln \sum {y_i} + ln {\frac{\sum -y_i}{\theta})}$$
We now take the partial derivtive, with respect to $\theta$
$$p´(y)= \frac {1}{\frac{n}{\theta^2}}*\frac {-2n}{\theta^3}+\frac{1}{ {\frac{\sum-y_i}{\theta}}}*\frac{\sum y_i}{\theta^2}$$
$$p´(y)=\frac{-2}{\theta}+\frac{1}{\theta}$$
Setting the derivative to 0 we get
$$\frac{-2}{\theta}=\frac{1}{\theta}$$
....which seems pretty disconcerting, if $\theta$ isn't approaching infinity.
The correct answer is supposed to be $\hat\theta =\frac{\bar y}{2}$
Where did I go astray?
Thankful for input/Magnus
Your logarithmized expression is not right. You can use, that $y_i=e^{\ln(y_i)}$
$$p(y)=\left( \frac{1}{\theta ^2} \right)^n\cdot \prod \limits_{i=1}^n e^{\ln(y_i)-\frac{y_i}{\theta}}$$
Taking ln
$$\ln(p)=n \cdot \ln \left( \frac{1}{\theta ^2} \right)+\sum \limits_{i=1}^n y_i-\frac{1}{\theta}\sum \limits_{i=1}^n y_i$$
Here you have to be careful. It is $\ln\left[\left( \frac{1}{\theta ^2} \right)^n\right]=n \cdot \ln \left( \frac{1}{\theta ^2} \right)$, not $\ln \left( \frac{n}{\theta ^2} \right)$
$\ln(p)=-n \cdot ln \left( \theta ^2 \right)+\sum \limits_{i=1}^n \ln(y_i)-\frac{1}{\theta}\sum \limits_{i=1}^n y_i$
$\ln(p)=-2n \cdot \ln \left( \theta \right)+\sum \limits_{i=1}^n \ln(y_i)-\frac{1}{\theta}\sum \limits_{i=1}^n y_i$
The derivative w.r.t $\theta$ is
$-2\frac{n}{\theta}+\frac{1}{\theta ^2} \sum \limits_{i=1}^n y_i=0$
Multiplying the equation by $\theta^2$
$-2n\theta+\sum \limits_{i=1}^n y_i=0$
I think you can take it from here.