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Jmath

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Hello I am trying to check if the Method of Moments and Maximum Likelihood Estimators for parameter $\theta$ from a sample with population density $$f(x;\theta) = \frac 2 \theta x e^{\frac {-x^2}{\theta}} $$

for $$x \geq 0$, $\theta > 0$$ with $\theta$ being unknown.

Taking the first moment of this function I found the Method of Moments estimator to be $$\hat{\theta}_1 = \frac{4\bar X^2}{\pi}$$ and solving for the Maximum Likelihood Estimator the Estimator to be $$\hat{\theta}_2 = 2\bar Y$$ where Y is just square of the Sample X_i, i.e. $$Y = X_i^2$$.

Steps in Solving for Method of Moments:

I took the first moment, i.e.

M_1 = E[x] = $$\int_0^\infty{\frac 2 \theta x^2 e^{\frac {-x^2}{\theta}}dx}$$

Solving this integral with $u$ substitution with $$u = \frac{-x}{2}, du = \frac{-1}{2}, v = e^\frac{x^2}{\theta}, dv = -2xe^\frac{-x^2}{\theta}$$

$$\int_0^\infty{\frac 2 \theta x^2 e^{\frac {-x^2}{\theta}}dx} = [-\frac{xe^\frac{-x^2}{\theta}}{2\theta} - \frac{\sqrt{\pi \theta}}{4}]^\infty_0 = \frac{\sqrt{\pi} {\sqrt{\theta}}}{2}$$

So that $$E[x] = \bar{x} = \frac{\sqrt{\pi} {\sqrt{\theta}}}{2}$ gives the Method of Moments Estimator $\hat{\theta_1} = \frac{4\bar{X}^2}{\pi}$$

Steps in Solving for Maximum Likelihood:

$$lnL(\theta)=(\prod_{i=1}^n\frac 2 \theta x e^{\frac {-x^2}{\theta}}) = -n ln((2\theta)) + \sum_{i=1}^nx_i - \frac {1} {\theta} \sum_{i=1}^nx^2_i$$

$$\frac {dlnL(\theta)}{d\theta} = \frac{-n}{2\theta} + \frac{1}{\theta^2} \sum_{i=1}^nx^2_i$$

Setting $\frac {dL(\theta)}{d\theta} = 0$, I found the Maximum Likelihood Estimator $\hat{\theta_2}$ to be $$\hat{\theta_2} = \frac{2\sum_{i=1}^nx^2_i}{n}$ , so that if $Y = X_i^2$ then $\hat{\theta_2} = 2\bar{Y}$$.

I am trying to check if these estimators for $\theta$ from this density function are unbiased and/or consistent but am lost on how to go about doing so, any help would be much appreciated.

for $$x \geq 0$, $\theta > 0$$ with $\theta$ being unknown.

Taking the first moment of this function I found the Method of Moments estimator to be $$\hat{\theta}_1 = \frac{4\bar X^2}{\pi}$$ and solving for the Maximum Likelihood Estimator the Estimator to be $$\hat{\theta}_2 = 2\bar Y$$ where Y is just square of the Sample X_i, i.e. $$Y = X_i^2$$.

Steps in Solving for Method of Moments:

I took the first moment, i.e.

M_1 = E[x] = $$\int_0^\infty{\frac 2 \theta x^2 e^{\frac {-x^2}{\theta}}dx}$$

Solving this integral with $u$ substitution with $$u = \frac{-x}{2}, du = \frac{-1}{2}, v = e^\frac{x^2}{\theta}, dv = -2xe^\frac{-x^2}{\theta}$$

$$\int_0^\infty{\frac 2 \theta x^2 e^{\frac {-x^2}{\theta}}dx} = [-\frac{xe^\frac{-x^2}{\theta}}{2\theta} - \frac{\sqrt{\pi \theta}}{4}]^\infty_0 = \frac{\sqrt{\pi} {\sqrt{\theta}}}{2}$$

So that $$E[x] = \bar{x} = \frac{\sqrt{\pi} {\sqrt{\theta}}}{2}$ gives the Method of Moments Estimator $\hat{\theta_1} = \frac{4\bar{X}^2}{\pi}$$

Steps in Solving for Maximum Likelihood:

$$lnL(\theta)=(\prod_{i=1}^n\frac 2 \theta x e^{\frac {-x^2}{\theta}}) = -n ln((2\theta)) + \sum_{i=1}^nx_i - \frac {1} {\theta} \sum_{i=1}^nx^2_i$$

$$\frac {dlnL(\theta)}{d\theta} = \frac{-n}{2\theta} + \frac{1}{\theta^2} \sum_{i=1}^nx^2_i$$

Setting $\frac {dL(\theta)}{d\theta} = 0$, I found the Maximum Likelihood Estimator $\hat{\theta_2}$ to be $$\hat{\theta_2} = \frac{2\sum_{i=1}^nx^2_i}{n}$ , so that if $Y = X_i^2$ then $\hat{\theta_2} = 2\bar{Y}$$.

I am trying to check if these estimators for $\theta$ from this density function are unbiased and/or consistent but am lost on how to go about doing so, any help would be much appreciated.

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