 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

Be careful about the index: the <probability density function> already contains a factor $x$, so $\mathbb E[X_i^r]=I_{r+1}$. The <recurrence relation> at $k=1$ yields $I_3=\theta I_1=\theta$. Therefore <linearity of expectation> gives
$$
\boxed{\mathbb E[\widehat\theta]=\frac1n\sum_{i=1}^n\mathbb E[X_i^2]=I_3=\theta.}
$$
Thus \b[$\widehat\theta$ is an <unbiased estimator>].

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