Show that $c_nX_n \overset{p}{\to}cX$ if $X_n \overset{p}{\to}X$ and $c_n$ is a sequence of reals which converges to the limit $c \in(0,\infty)$.

Question

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.

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Minkowski's inequality:

$(E(|X+Y|^r))^{1/r}\le (E(|X|^r))^{1/r}+(E(|Y|))^{1/r}$

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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?

Answer 1

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 $\frac {\epsilon} {|c_n-c|} \to \infty$..