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## 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]