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I am not sure how to do this. To find the method of moments estimator I did:

$$E[X] = \frac{-\theta + \theta}{2} = 0$$

use 2nd moment:

$$E[X^2] = \frac{(-\theta)^2 + -(\theta^2) + \theta^2}{3} = \frac{\theta^2}{3} = \frac{1}{n} \sum_{i=1}^n X_i^2$$

$$\hat{\theta} = \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2}$$

And to find if it is unbiased, I tried this, but am not sure what to do with the integral

$$\begin{align}E[\hat{\theta}] &= \sqrt{\frac{3}{n}} \mathbb{E}\left[\sqrt{\sum_{i=1}^n X_i^2} \right] \\ &= \int_{-\theta}^\theta \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2} \frac{1}{2\theta} dx \\ &= \sqrt{\frac{3}{n}} * \frac{1}{2\theta} \int_{-\theta}^\theta \sqrt{\sum_{i=1}^n X_i^2} dx \end{align}$$

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    $\begingroup$ Observe that $(3/n)\sum X_i^2$ is an unbiased estimator of $\theta^2$. What does Jensen's Inequality tell us about the relationship between the expected value of a square root and the square root of the expected value? $\endgroup$
    – jbowman
    Commented Mar 1 at 1:55

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If we apply Jensen's Inequality as @jbowman suggested, which states that for a concave function $f$ (in this case, the square root function), and a random variable $Y$, we have: $$E[f(Y)] \leq f(E[Y]).$$ Therefore, since $\hat{\theta}$ involves taking a square root (a concave function), we have: $$E[\hat{\theta}] = E\left[\sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2}\right] < \sqrt{E\left[\frac{3}{n} \sum_{i=1}^n X_i^2\right]} = \sqrt{3E[X^2]} = \sqrt{\theta^2} = \theta.$$

Since $E[\hat{\theta}]$ is less than $\theta$, we conclude that $\hat{\theta}$ is a biased estimator for $\theta$. Specifically, the bias is $$Bias(\hat{\theta}) = E[\hat{\theta}] - \theta < 0,$$ showing that the estimator tends to underestimate the true parameter.

Update: The transition from a weak inequality to a strict one in the application of Jensen's Inequality for the estimator $\hat{\theta} = \sqrt{\frac{3}{n} \sum_{i=1}^n X_i^2}$ occurs due to the strict concavity of the square root function. Jensen's Inequality tells us that for a concave function $f$ and a random variable $Y$, $E[f(Y)] \leq f(E[Y])$. This inequality is typically strict (i.e., $E[f(Y)] < f(E[Y])$) unless $Y$ is constant almost surely , which is not the case for sums of squared random variables unless they are all the same value almost surely.

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    $\begingroup$ You might want to mention why the weak inequality becomes strict all of a sudden $\endgroup$
    – Taylor
    Commented Mar 1 at 4:10

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