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I'm working on a problem from my mathematical statistics book and the following is asked from me. Let $X_{1},...,X_{n}$ be a sample from a probability density function with density $p_{\theta}(x)= \frac{2x}{\theta^{2}}1_{0 \leq x \leq \theta}.$

1) Determine method of moments estimator for $\theta$.

2) Determine method of moments estimator for $\theta^{2}$.

I think I've found the solution for 1, but I have no idea where to start with the solution for 2. Any hints would be much appreciated. So far my solution for 1)

Because we are determining a method of moments estimator for $\theta$, we set $E(X_{i}^{j})=\bar{X^{j}}$. In this case we let $j=1$, since that solution exists as we shall see.

We define a moments estimator $T$ for $\theta$ as follows, \begin{equation} \begin{split} E(p_{\theta}) &= \int_{-\infty}^{\infty} x p_{\theta}(x) dx\\ &= \frac{2}{\theta^2} \int_{0}^{\theta} x^{2} dx \text{ (since $1_{0 \leq x \leq \theta}$ we let $0$ and $\theta$ be the boundaries for $x$)} \\ &= \frac{2}{\theta^2} \left[ \frac{1}{3}x^{3} \right]_{x=0}^{x=\theta}\\ &= \frac{2\theta}{3} = \bar{X}. \end{split} \end{equation} Which implies $\hat{\theta} = \frac{3\bar{X}}{2}$ is an estimator for $\theta$.

But now, how to determine an estimator for $\theta^{2}$? Do I just substitute $\theta^{2}$ for $\theta$ so $p_{\theta^{2}}=\frac{2x}{\theta^{4}}$ and then iterate my previous method?

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2 Answers 2

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Just as you did with Q1, in the same manner, you could say \begin{equation} E(\theta^2) = \frac{2}{\theta^2} \int_0^{\theta} x^3 \ dx = \frac{2}{\theta^2} \frac{1}{4} \left[ x^4\right]_{x=0}^{x=\theta} = \frac{\theta^2}{2} = \bar{X}_2 \end{equation} where $\bar{X}_2 = \frac{1}{n}\sum_{i=1}^n x_i^2$ is the sample estimate of the second moment. So you could estimate $\theta$ as \begin{equation} \hat{\theta} = \sqrt{2\bar{X}_2} \end{equation}

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  • $\begingroup$ Thanks for the fast reply! I already had this written down but wasn't sure about the method. Thanks for clarifying! $\endgroup$ Sep 25, 2018 at 15:34
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Do I just substitute $\theta^{2}$ for $\theta$ so $p_{\theta^{2}}=\frac{2x}{\theta^{4}}$ and then iterate my previous method?

No, this only applies for maximum likelihood estimation, and not method of moments estimation.

For the method of moments, start by computing the second moment: $$\mathbb{E}[X^2] = \dfrac{2}{\theta^2}\int_{0}^{\theta}x^3\text{ d}x = \dfrac{2}{\theta^2} \cdot \dfrac{\theta^4}{4} = \dfrac{\theta^2}{2}\text{.}$$ By the method of moments, $$\dfrac{\hat{\theta^2}}{2}=\dfrac{1}{n}\sum_{i=1}^{n}X_i^2$$ hence $$\hat{\theta^2} = \dfrac{2}{n}\sum_{i=1}^{n}X_i^2\text{.}$$

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  • $\begingroup$ which is equal to $2 \bar{X^{2}}$, right?And thanks for the solution! $\endgroup$ Sep 25, 2018 at 15:32
  • $\begingroup$ @Mathbeginner $\bar{X}^2$ is not the same as $\dfrac{1}{n}\sum_{i=1}^{n}X_i^2$. Note that $\bar{X}^2 = \left(\dfrac{1}{n}\sum_{i=1}^{n}X_i \right)^2 = \dfrac{1}{n^2}\left(\sum_{i=1}^{n}X_i \right)^2$, which is not the same as what I wrote above. $\endgroup$ Sep 25, 2018 at 15:41
  • $\begingroup$ You're totally right, thanks! $\endgroup$ Sep 25, 2018 at 15:46

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