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

Let Xi be an iid random variable having pdf f(x|θ), where E(Xi)=6θ2, and θ>0.

I have calculated an estimator for the parameter (θ) of f(x|θ) to be ˆθ=ˉx/6. To prove that this is an unbiased estimator, I should prove that E(ˆθ)=E(ˉx/6). However, since ˆθ2=ˉx/6, it would be much easier to show that E(ˆθ2)=E(ˉx/6)=16E(Xin)=16nE(Xi)=16nn6θ2=θ2.

Generally, proving x2=4 is not the same as proving x=2, since x could also be 2. However, in this case θ>0.

I have shown that ˆθ2 is unbiased, is this sufficient to show that ˆθ is unbiased?

Answer

Say Q is unbiased for \theta^2, i.e. E(Q) = \theta^2, then because of Jensen’s inequality,

\sqrt{ E(Q) } = \theta < E \left( \sqrt{Q} \right)

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.

Attribution
Source : Link , Question Author : kingledion , Answer Author : gammer

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