# Thread: help with convergence in probability

1. ## help with convergence in probability

If Xn->X (in probability) and Yn->Y(in probability) and Cn=Xn+Yn and C=X+Y show that Cn->C (in probability).

what i got so far
|Xn-X|<=ε/2 and |Yn-Y|<=ε/2
|(Xn+Yn)-(X+Y)|=|Xn-X+Yn-Y|<=|Xn-X|+|Yn-Y|<=ε/2+ε/2=ε

|(Xn+Yn)-(X+Y)|<=ε
|Cn-C|<=ε
so lim(n->inf) P(|Cn-C|>=ε)=0

I want to verify if this is correct, and if not, what's the mistake, or if I am missing anything.

Thanks

2. The convergence in probability does not say $\displaystyle |X_n-X|<\epsilon$, what you have is for each epsilon>0, $\displaystyle \mathbb{P}(|X_n-X|>\epsilon)\rightarrow 0$ as n increases.

The way you approach this problem is in terms of sets. What can you say about $\displaystyle \{\omega:|X_n(\omega)+Y_n(\omega)-X(\omega)-Y(\omega)|>\epsilon\}$ in terms of $\displaystyle \{\omega: |X_n(\omega)-X(\omega)|> \epsilon/2\}\cup\{\omega: |Y_n(\omega)-Y(\omega)|> \epsilon/2\}$?

3. {w:|(Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε}
{w:|Cn(w)-C(w)|>=ε}
P(|Cn-C|>=ε)->0
so lim(n->inf) P(|Cn-C|>=ε)=0

Is that right?

4. Originally Posted by Sneaky
{w:|(Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε}
{w:|Cn(w)-C(w)|>=ε}
P(|Cn-C|>=ε)->0
so lim(n->inf) P(|Cn-C|>=ε)=0

Is that right?
It does converge but why? Why do you have the third line? You want to bound $\displaystyle \mathbb{P}(|C_n-C|>\epsilon)$. Read what I have written before and use $\displaystyle \mathbb{P}(A) \leq \mathbb{P}(B)$ whenever $\displaystyle A \subset B$.

5. {w:|(Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε}
{w:|Cn(w)-C(w)|>=ε}

so

{w:|(Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε}
contains
{w:|Xn(w)+X(w)|>=ε/2}U{w:|Yn(w)+Y(w)|>=ε/2}

so

P(|Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε) >= P(|Xn(w)+X(w)|>=ε/2)
P(|Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε) >= P(|Yn(w)+Y(w)|>=ε/2)

P(|Cn-C|>=ε) >= P(|Xn(w)+X(w)|>=ε/2)
P(|Cn-C|>=ε) >= P(|Yn(w)+Y(w)|>=ε/2)

Am I on the right track here?

6. Originally Posted by Sneaky
{w:|(Xn(w)+Yn(w))-(X(w)+Y(w))|>=ε}
contains
{w:|Xn(w)+X(w)|>=ε/2}U{w:|Yn(w)+Y(w)|>=ε/2}
If you have two events such that "If A, then B", that tells you $\displaystyle A \subset B$.