Gregg
- 452
- 0
Homework Statement
[itex]X, Y, (X_n)_{n>0} \text{ and } (Y_n)_{n>0}[/itex] are random variables.
Show that if
[itex]X_n \xrightarrow{\text{P}} X[/itex] and [tex]Y_n \xrightarrow{\text{P}} Y[/tex] then [itex]X_n + Y_n \xrightarrow{\text{P}} X + Y[/itex]
Homework Equations
If [itex]X_n \xrightarrow{\text{P}} X[/itex] then [itex]\text{Pr}(|X_n-X|>\epsilon)=0 \text{ } \forall \epsilon > 0 \text{ as } n \to \infty[/itex]
The Attempt at a Solution
First, let the sets [itex]A_n(\epsilon) = \{|X_n - X|<\epsilon\}[/itex] and [itex]B_n(\epsilon) = \{|Y_n - Y|<\epsilon\}[/itex]
The sum of the two moduli will always be less than [itex]2\epsilon[/itex] if both of the moduli are less than [itex]\epsilon[/itex] but the converse is not generally true.
[itex]C_n(\epsilon)=\{|X_n-X|+|Y_n-Y|<2\epsilon\}\supset{A_n(\epsilon)\cap B_n(\epsilon)}[/itex]
Using the triangle inequality:
[itex]|X_n + Y_n - X - Y | \le |X_n-X|+|Y_n-Y|[/itex]
[itex]D_n(\epsilon) =\{|X_n-X+Y_n-Y|<2\epsilon\} \supset C_n(\epsilon)[/itex]
I think this has gone wrong in several places but from here I hope to say that
[itex]\text{Pr}(D_n) \ge \text{Pr}(C_n) \ge \text{Pr}(A_n\cap B_n ) \ge \text{Pr}(A_n) \to 1 \text{ as } n\to\infty[/itex]
[tex]\text{Pr}(D_n^c) \to 0 \text{ as } n\to \infty[/tex]