Suppose Xn P-> X and a is a constant. Then aXn P-> aX ?

QUESTION:

Suppose $X_{n}\overset{P}{\rightarrow}X$ and a is a constant. Then $aX_{n}\overset{P}{\rightarrow}aX$ ?

SOLUTION:

First, the result holds trivially if $a=0$ so we can suppose without loss of generality that $a\neq 0$. We have

$P\left ( \left | aX_{n}-aX \right |\geq \varepsilon  \right )=P\left ( \left | a \right |\left | X-X_{n} \right |\geq \varepsilon  \right )=P\left ( \left | X_{n} -X\right |\geq \frac{\varepsilon }{\left | a \right |}  \right )$

Since  $X_{n}\overset{P}{\rightarrow}X$ the $\lim_{n \to \infty }P\left ( \left | X_{n} -X\right |\geq \frac{\varepsilon }{\left | a \right |}  \right )=0$  so (by the Sandwich Theorem, say)  $\lim_{n \to \infty }P\left ( \left | aX_{n} -aX\right |\geq \varepsilon  \right )=0$  so that  $aX_{n}\overset{P}{\rightarrow}aX$, as claimed.