# 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 $c \in(0,\infty)$.

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?

• Are you sure that you stated the problem and the hint correctly? In general, $X$ doesn't need to be integrable, and so it doesn't make sense to approach the problem this way. – saz Sep 23 '18 at 8:55
• Yes, I checked several times. However, I did forget to write that $n\to\infty$. I've corrected that above. – AnnieFrannie Sep 23 '18 at 9:31
• Ok, so how should I approach the problem then? – AnnieFrannie Sep 23 '18 at 9:46
• Take a look at this question: math.stackexchange.com/q/1544449/36150 ... it is somewhat more general than yours; simply set $Y_n :=c_n$. – saz Sep 23 '18 at 10:27

## 1 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 $$c_n . Then $$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$$..