# Convergence in probability in relation to large sample variance

I have a (possibly simple) question:

If a sequence of scalar random variables, $\{ X_T \}_{T=1}^{\infty}$, convergences in probability to a constant $c$, does that imply that the variance of $X_T$ converges to $0$? In other words, if one has an estimator $\hat{\beta}_T$ of a true parameter value $\beta_0$ with $Var(\hat{\beta}_T)$ converging to a non-zero constant, can this estimator be a consistent estimator of $\beta_0$?

-
It's worth mentioning that although the answer to this question is "no," as in the answers below, there do exist conditions such that convergence in probability does guarantee that the variance converges to 0; if the $X_t^2$ are uniformly integrable you get the implication, but depending on your level this might be quite technical. –  guy May 11 '12 at 17:43

No. Try $\mathrm P(X_T=0)=1-2/T$ and $\mathrm P(X_T=2^T)=\mathrm P(X_T=-2^T)=1/T$.

Edit Consider an estimator $X_T$ of a true value $c$ and assume that $X_T$ converges to $c$ in probability, thus $X_T$ is consistent. Then, the variance of $X_T$ may or may not go to zero. In particular the variance may not go to zero and the estimator be nevertheless consistent.

-
Ok. Then, does the following state make sense?: "We will prove that in general $\hat{\beta}_T$ is not a consistent estimator of $\beta_0$ be showing that its variance converges to a non-zero value as to sample size increases to $\infty$" –  Esben Vibel May 11 '12 at 12:02
No. See Edit.  –  Did May 11 '12 at 15:39

No. Another example: imagine your parameter $\hat{\beta}_T$ happens to follow a Cauchy distribution with fixed center and scale ("width") that tends to zero with $T$. Then, it will converge in probability, but its variance will no tend to zero - actually, it will be infinite for all $T$.

Often one proves consistency by showing that the MSE (or both variance and bias) tends to zero. But this is a sufficient, not necessary condition. So one cannot prove that an estimator is not-consistent by showing that its variance does not tend to zero: that in itself proves nothing.

-
Ok. Thanks a lot! –  Esben Vibel May 11 '12 at 13:16
@leonbloy It's a good answer, but I'd prefer to say that "often one proves consistency by showing that the MSE tends to zero". We also need to worry about bias. –  Byron Schmuland May 11 '12 at 15:43
@ByronSchmuland: good observation, fixed. –  leonbloy May 11 '12 at 15:46