I'm new at this and need help with the following problem:
"Assume $X_n \overset{p}{\to}X$ and $c_n$ is a sequence of reals which converges to the limit $c \in(0,\infty)$. Show that $c_nX_n \overset{p}{\to}cX$."
I got a hint that says that I can start out, as $n \to \infty$, with
$E(|c_nX_n-cX|^r)\le |c_n|^rE(|X_n-X|^r)+|c_n-c|^rE|X^r| \to0$,
by Minkowski's inequality.
Minkowski's inequality:
$(E(|X+Y|^r))^{1/r}\le (E(|X|^r))^{1/r}+(E(|Y|))^{1/r}$
The hint confuses me, why is it ok to just remove the brackets $(\cdot)^{1/r}$ and where does the $c_n$ come from (in the term $|c_n-c|^rE|X^r|$)? Why does the right hand side of this converge to zero?
Best Answer
As pointed out by saz the hint is bad. You can prove this from definition of convergence in probability as follows :choose $n$ such that $c_n <c+1$. Then $P\{|c_nX_n-cX|>\epsilon\} \leq P\{|c_n(X_n-X)|>\epsilon /2\}+P\{|(c_n-c)X|>\epsilon/2\}$. First term is $\leq P\{|X_n-X|>\epsilon /{2c_n}\} \leq P\{|X_n-X|>\epsilon /{2(c+1)}\} \to 0$. Second term is $\leq P\{|X|>\frac {\epsilon} {|c_n-c|}\} \to 0$ because $\frac {\epsilon} {|c_n-c|} \to \infty$..