Maximum likelihood estimator

[semidevil]

Maximum likelihood estimator...

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 doesn't 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:

$$fy(y;\theta) = \frac{y^3e^{\frac{-y}{\theta}}}{6\theta^4}$$

 

[Valkarie]

The maximum likelihood estimator for θ is given by solving the equation: \frac {1}{6n\theta^4} -\frac{ 3 * \sum_1^n yi^2}{\theta^5} e^{\frac{-y_i}{\theta}} = 0.Can someone help me out?

 

[tacman]

First, let's clarify the notation a bit. The function is actually f(y;θ), where y is the random variable and θ is the parameter we are trying to estimate. So the first step is to write out the likelihood function:

L(θ) = \prod_{i=1}^n f(y_i;θ) = \prod_{i=1}^n \frac{2y_i}{θ^2} = \frac{2^n}{θ^{2n}} \prod_{i=1}^n y_i

Next, we take the natural log of the likelihood function:

ln L(θ) = nln2 - 2nlnθ + \sum_{i=1}^n ln y_i

To find the maximum likelihood estimator, we set the derivative of ln L(θ) with respect to θ equal to 0 and solve for θ:

\frac{d}{dθ} ln L(θ) = -\frac{2n}{θ} + \sum_{i=1}^n \frac{1}{y_i} = 0

Solving for θ, we get:

\hat{θ}_{MLE} = \frac{1}{n} \sum_{i=1}^n y_i = \bar{y}

So the maximum likelihood estimator for θ is simply the sample mean.

Now for the method of moments, we set the first moment of the distribution (E[y]) equal to the first moment of the sample (\bar{y}) and solve for θ:

E[y] = \int_0^θ y \frac{2y}{θ^2} dy = \frac{2}{θ^2} \int_0^θ y^2 dy = \frac{2}{θ^2} \frac{θ^3}{3} = \frac{2}{3}θ

Setting this equal to the sample mean, we get:

\frac{2}{3}θ = \bar{y}

Solving for θ, we get:

\hat{θ}_{MoM} = \frac{3}{2} \bar{y}

So the method of moments estimator for θ is 1.5 times the sample mean.