Variance of $X_n$ if we have convergence in probability for a random variable with $|X_n| >1$ Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$.  Assume $X_n  \overset{p}{\to}  X$.  True or false: $Var(X_n)\to 0$.  If false, give a counterexample.  If true, give an argument.
Convergence of $\operatorname{Var}(X_n)$ if $|X_n| \le 1$ seems relevant. How much does the $X_n \to 0$ change what I have to say? Would appreciate a start in the right direction thanks!
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Suppose $X_n$ is a sequence of random variables with $|X_n| \leq 1$.  Assume $X_n  \overset{p}{\to}  X$.  True or false: $Var(X_n)\to 0$.  If false, give a counterexample.  If true, give an argument. \
\indent The statement is false. Here we can offer a counter example.\
\
\indent Let $X \sim U[0,1]$. Since $X_n  \overset{p}{\to}  X$ we know $X_n  \overset{d}{\to}  X$. This implies that $Var(X_n) = Var(X)$. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can then follow the steps outlined here and say:
\begin{equation*}
\begin{aligned}
& \operatorname{Var}(X) = \int_{-\infty}^\infty x^2 f_X(x) \, dx - \left( E(X) \right)^2 \\
& = \int_{-\infty}^0 0 f_X(x) \, dx + \int_0^1 x^2 f_X(x) \, dx + \int_0^\infty 0 f_X(x) \, dx - \left( \frac{0+1}{2} \right)^2 \\
& = \int_0^1 x^2 (1) \, dx - \frac{1}{4} \\
& = \left( \frac{x^3}{3} \right) \bigg\rvert_0^1 \\
& = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\
& = \frac{1}{12}
\end{aligned}
\end{equation*}\
 A: For your first question - obviously false.  Let $X_m=X_n$ for all indices, so they converge trivially to the same random variable with the same non-zero variance.
I suspect that for the second question, the variance would converge to the variance of the limit, but I haven't tried to work it out.
A: I added an edit that posted my answer after following some advice given! Thanks to Teresa!
Let $X \sim U[0,1]$. Since $X_n  \overset{p}{\to}  X$ we know $X_n  \overset{d}{\to}  X$. This implies that $X_n \sim U[0,1]$ $Var(X_n) = Var(X)$ and we will show this $\neq 0$ for at least the following case. Using the definition of the continuous uniform distribution we can say that $f_X(x) = \frac{1}{1-0} = \frac{1}{1}$. We can say:
\begin{equation*}
\begin{aligned}
& Var(X) = \int_{-\infty}^{\infty} x^2 f_X(x) dx - \left( E(X) \right)^2 \\
& = \int_{-\infty}^{0} 0 f_X(x) dx + \int_{0}^{1} x^2 f_X(x) dx + \int_{0}^{\infty} 0 f_X(x) dx - \left( \frac{0+1}{2} \right)^2 \\
& = \int_{0}^{1} x^2 (1) dx - \frac{1}{4} \\
& = \left( \frac{x^3}{3} \right) \bigg\rvert_{0}^{1} \\
& = \left( \frac{1}{3} - \frac{0}{3} \right) - \frac{1}{4} \\
& = \frac{1}{12}
\end{aligned}
\end{equation*}\
