# If I prove the estimator of θ2\theta^2 is unbiased, does that prove that the estimator of parameter θ\theta is unbiased?

Let $X_i$ be an iid random variable having pdf $f(\mathbf{x}|\theta)$, where $E(X_i) = 6\theta^2$, and $\theta > 0$.

I have calculated an estimator for the parameter ($\theta$) of $f(\mathbf{x}|\theta)$ to be $\hat{\theta} = \sqrt{\bar{x}/6}$. To prove that this is an unbiased estimator, I should prove that $E(\hat{\theta}) = E\left(\sqrt{\bar{x}/6}\right)$. However, since $\hat{\theta}^2 = \bar{x}/6$, it would be much easier to show that

Generally, proving $x^2 =4$ is not the same as proving $x=2$, since $x$ could also be $-2$. However, in this case $\theta>0$.

I have shown that $\hat{\theta}^2$ is unbiased, is this sufficient to show that $\hat{\theta}$ is unbiased?

Say $Q$ is unbiased for $\theta^2$, i.e. $E(Q) = \theta^2$, then because of Jensen’s inequality,
So $\sqrt{Q}$ is biased high, i.e. it will overestimate $\theta$ on average.
Note: This is a strict inequality (i.e. $<$ not $\leq$) because $Q$ is not a degenerate random variable and square root is not an affine transformation.