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Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply variance approaching zero.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=-x^{-2}\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Edit: If you want an example in which the variance is finite, take $$\xi_n(x):=\chi_{[0,1/n]}(x)\sqrt{n},$$ and again consider $\eta_n:=\pm\xi_n$ w.p. 1/2.

Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply variance approaching zero.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=\log(x)\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Edit: If you want an example in which the variance is finite, take $$\xi_n(x):=\chi_{[0,1/n]}(x)\sqrt{n},$$ and again consider $\eta_n:=\pm\xi_n$ w.p. 1/2.

Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply variance approaching zero.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=-x^{-2}\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Edit: If you want an example in which the variance is finite, take $$\xi_n(x):=\chi_{[0,1/n]}(x)\sqrt{n}.$$

Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply variance approaching zero.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=-x^{-2}\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Edit: If you want an example in which the variance is finite, take $$\xi_n(x):=\chi_{[0,1/n]}(x)\sqrt{n},$$ and again consider $\eta_n:=\pm\xi_n$ w.p. 1/2.

Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply finite variance.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=-x^{-2}\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Let me give an example of a sequence of random variable converging to zero in probability but with infinite variance. In essence, an estimator is just a random variable so with a little abstraction, you can see that convergence in probability to a constant does not imply variance approaching zero.

Consider the random variable $\xi_n(x):=\chi_{[0,1/n]}(x)x^{-1/2}$ on $[0,1]$ where the probability measure considered is the Lebesgue measure. Clearly, $P(\xi_n(x)>0)=1/n\to0$ but $$\int\xi_n^2dP=\int_{0}^{1/n}x^{-1}dx=-x^{-2}\mid _{0}^{1/n}=\infty,$$ for all $n$ so its variance does not go to zero.

Now, just make up an estimator where as your sample grows you estimate the true value $\mu=0$ by a draw of $\xi_n$. Note that this estimator is not unbiased for 0, but to make it unbiased you can just set $\eta_n:=\pm\xi_n$ with equal probability 1/2 and use that as your estimator. The same argument for convergence and variance clearly holds.

Edit: If you want an example in which the variance is finite, take $$\xi_n(x):=\chi_{[0,1/n]}(x)\sqrt{n}.$$