semidevil
Feb22-05, 05:20 PM
ok, i'm stil a bit lost...so tell me if this is right:
f_y(y;\theta) = \frac{2y}{\theta^2}, for 0 < y < \theta
find the MLE estimator for theta.
L(\theta) = 2yn\theta^{-2 \sum_1^n y_i .
is this even right to begin with?
then take the natural log
ln2yn + -2\sum_1^n y_i * ln\theta
take derivative
\frac {1}{2yn} -\frac{ 2 * \sum_1^n yi}{\theta}.
now how do I solve this in terms of theta? and after that, what do I do next?
this just doesnt look right
and I also need to find it using the method of moments, but I get \frac {y^4}{2\theta} after the integral...
and this one too...this looks too messy:
fy(y;\theta) = \frac{y^3e^{\frac{-y}{\theta}}}{6\theta^4}
f_y(y;\theta) = \frac{2y}{\theta^2}, for 0 < y < \theta
find the MLE estimator for theta.
L(\theta) = 2yn\theta^{-2 \sum_1^n y_i .
is this even right to begin with?
then take the natural log
ln2yn + -2\sum_1^n y_i * ln\theta
take derivative
\frac {1}{2yn} -\frac{ 2 * \sum_1^n yi}{\theta}.
now how do I solve this in terms of theta? and after that, what do I do next?
this just doesnt look right
and I also need to find it using the method of moments, but I get \frac {y^4}{2\theta} after the integral...
and this one too...this looks too messy:
fy(y;\theta) = \frac{y^3e^{\frac{-y}{\theta}}}{6\theta^4}