1

A sequence $\{X_n\}$ of random variables converges in probability towards the random variable $X$, Similarly, $\{Y_n\}$ converges in probability to $Y$. Under which conditions can I say that $\{X_n - Y_n\}$ converges to $X-Y$? Is there a theorem giving sufficient condition for this? Given that the examples of cases where this is not true seem very knife-edge.

fox
  • 521
  • 1
    A suitable $\epsilon, \delta, N_0$ argument, perhaps using $\frac{\delta}{2}$, should lead to a proof – Henry Feb 24 '18 at 23:13

1 Answers1

3

Fix $\varepsilon,\delta>0$. By assumption there exists $N_0$ big enough so that $P(|X-X_n|>\varepsilon/2)<\delta/2$ and $P(|Y-Y_n|>\varepsilon/2)<\delta/2$ for all $n>N_0$. Then, for all $n>N_0$ \begin{align*}P(|X-Y-(X_n-Y_n)|>\varepsilon)&\leq P(|X-X_n|+|Y-Y_n|>\varepsilon)\\ &\leq P(|X-X_n|>\varepsilon/2)+P(|Y-Y_n|>\varepsilon/2)<\delta. \end{align*}

K.Power
  • 7,111
  • So this argument shows that this is true for every X and every Y? – fox Feb 25 '18 at 15:08
  • Unless I've made a terrible mistake somewhere then yes. More generally the continuous mapping theorem ensures the result definitely holds for any $Y_n\to Y$ and $X_n\to X$. – K.Power Feb 25 '18 at 15:23
  • 2
    If $X_n \overset{p}{\to} X $ and $Y_n \overset{p}{\to} Y$ then you can prove that $(X_n, Y_n) \overset{p}{\to} (X, Y )$. There is a theorem (Continuous mapping theorem which says that if $f$ is a continuous function and $(X_n, Y_n) \overset{p}{\to} (X, Y)$ then $f(X_n, Y_n) \overset{p}{\to} f(X, Y)$. In particular the function f(x,y)= x-y is a continuous function. – Joogs Feb 25 '18 at 15:25