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?

  • $\begingroup$ 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. $\endgroup$ – saz Sep 23 '18 at 8:55
  • $\begingroup$ Yes, I checked several times. However, I did forget to write that $n\to\infty$. I've corrected that above. $\endgroup$ – AnnieFrannie Sep 23 '18 at 9:31
  • $\begingroup$ Ok, so how should I approach the problem then? $\endgroup$ – AnnieFrannie Sep 23 '18 at 9:46
  • $\begingroup$ Take a look at this question: math.stackexchange.com/q/1544449/36150 ... it is somewhat more general than yours; simply set $Y_n :=c_n$. $\endgroup$ – saz Sep 23 '18 at 10:27

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


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