Let \(X_n\sim N(0, 1+1/n)\) and \(X\sim N(0,1)\). Find a joint distribution of \((X_n,X)\) so that \(X_n\rightarrow X\) in probability.
Say\((X_n,X)\) is bivariate normal with \(\mu_{X_n}=\mu_X=0\), \(\sigma_{X_n}=1+1/n\), \(\sigma_X=1\) and \(cor(X_n,X)=1\), then \(X_n-X\) has a normal distribution with
\[ \begin{aligned} &E[X_n-X] = 0\\ &var(X_n-X) =var(X_n)+var(X)-2cov(X_n,X)=\\ &(1+1/n)^2+1^2-2(1+1/n)\times 1\times 1 =\\ &1+2/n+1/n^2+1-2-2/n = 1/n^2 \end{aligned} \] and so
\[ \begin{aligned} &P(|X_n-X|<\epsilon) = \\ &2F_{X_n-X}(\epsilon)-1 =\\&2\Phi(n\epsilon)-1 \rightarrow 2\times 1-1=1 \end{aligned} \] as \(n\rightarrow \infty\) because then \(n\epsilon \rightarrow \infty\).
Let \(X\sim U[0,1]\) and \(X_n|X=x\sim U[x,x+1/n]\). Show that \(X_n\rightarrow X\) in probability.
We have \(X<X_n<X+1/n\), or \(0<X_n-X<1/n\). Now let \(n_\epsilon=\lceil 1/\epsilon\rceil\), so that for all \(n>n_\epsilon\) we have \(\epsilon>1/n\) and therefore
\[ \begin{aligned} &P(|X_n-X|<\epsilon) = \\ &P(X_n-X<\epsilon) \ge \\ &P(X_n-X<1/n) = 1 \end{aligned} \]
Let \(X_n\) be a random variable with \(P(X_n=0)=1/n,P(X_n=1)=1-2/n,P(X_n=2)=1/n\). Show that \(X_n\rightarrow 1\) in probability
if \(\epsilon<1\) we have
\[ \begin{aligned} &P(|X_n-X|<\epsilon) = P(|X_n-1|<\epsilon)= \\ &P(X_n=1) = 1- 2/n \rightarrow 1\\ \end{aligned} \]
\[ \begin{aligned} &E[X] = 0\times 1/n + 1\times (1-2/n) +2\times 1/n = 1\\ &E[X^2] = 0^2\times 1/n + 1^2\times (1-2/n) +2^2\times 1/n = 1+2/n\\ &var(X_n) = 1+2/n-1^2=2/n\rightarrow 0\\ \end{aligned} \] so \(X_n\rightarrow 1\) in quadratic mean, and therefore by theorem 3.2.9a also in probability.