# A little help with a beginning stats proof

1. Jun 7, 2010

### bennyska

1. The problem statement, all variables and given/known data

Let A_1, A_2... be any infinite sequence of events, and let B_1, B_2... be another infinite sequence defined as B_1=A_1, B_2=A_1cA_2, B_3=A_1cA_2cA_3 and so on.
Prove that Pr(Union i=1 to n A_i) = Sum i=1 to n Pr(B_i).
(sorry if that notation is hard to understand)

2. Relevant equations

3. The attempt at a solution

so i'm somewhat new to statistics proofs, but this one is for the most part a sets proof, which i can do. i'm having trouble connecting them.
(c = complement)(AB = A intersect B)
)

So i've convinced myself that this is true. I see if i take the union of A_i up to n, that B_i up to n is equal. Each B_i is A_i minus any previous As that intersect it. I'm just having trouble saying that in math. If I had to write a proof right now, I'd say each Sum B_i = Union A_i, so they're equal.

So yeah, if i could get some hints on where to start. Thanks.

2. Jun 7, 2010

### Tedjn

Is it the union from i = 1 to n or the union from i = 1 to infinity?

3. Jun 8, 2010

### bennyska

i=1 to n, although the next proof is the same problem with i=1 to infinity. i'm hoping once i get this proof down that should be a simple extension.

4. Jun 8, 2010

### Tedjn

If you accept that probabilities are countably additive, both are easy. What facts are you allowed to assume?

5. Jun 8, 2010

### lanedance

so do you mean $B_2=A_1^c \cap A_2$?

if so you should be able to show the sets are equivalent & the probabilities will follow...

6. Jun 8, 2010

### Staff: Mentor

Math induction seems to me that way to go. The base case is trivial, since A1 = B1.

Assume that the statement is true for n = k - 1, and show why the statement must be true for n = k.

The statement for n = k - 1 is
$$Pr\left( \bigcup_{i = 1}^{k - 1} A_i \right)= \sum_{i = 1}^{k - 1} Pr(B_i)$$

7. Jun 9, 2010

### bennyska

excuse the sloppy notation, and AB means A intersect B:

Let x be in U(i=1 to n)A_i. Then x is in A_i for some i<=n in I. Then x is in U(i=i to n)B_i, since U B_i = A_1 + A_1cA_2 + A_1cA_2cA_3 + ... + ...A_n-1cA_n = A_1UA_2U...UA_n, so U(i=1 to n)A_i is a subset of U(i=1 to n)B_i.
Let y be in U(i=1 to n)B_i. Then y is in A_i for some i<=n in I. So y is in U(i=1 to n)A_i, and U(i=1 to n)B_i is a subset of U(i=1 to n)A_i, and the two sets are equal.
Since B_1, B_2,...B_n are disjoint events, then Sum(i=1 to n)Pr(B_i) = Pr(U(i=1 to n)B_i), and since B_i = A_i, then Pr(U(i=1 to n)A_i) = Pr(U(i=1 to n)B_i).

how does that look (aside from the sloppy notation?). Is the proof more or less the same for the next proof, which is the same, except it is from i=1 to infinity?

8. Jun 9, 2010

### EnumaElish

Using "+" between sets is not standard. You can add probabilities but not sets.

$$Pr\left( \bigcup_{i = 1}^{k - 1} A_i \right)= \sum_{i = 1}^{k - 1} Pr(B_i)$$

is true. Now write:

$$Pr\left( \bigcup_{i = 1}^{k} A_i \right) = Pr\left(\bigcup_{i = 1}^{k - 1} A_i \cup A_k\right)$$
and
$$\sum_{i = 1}^{k} Pr(B_i) = \sum_{i = 1}^{k-1} Pr(B_i) + Pr(B_k)$$
and show the two are equal, using
$$Pr\left( \bigcup_{i = 1}^{k - 1} A_i \right)= \sum_{i = 1}^{k - 1} Pr(B_i)$$

Last edited: Jun 9, 2010
9. Jun 10, 2010

### bennyska

so i'm having a hard time getting latex to work right (i figured out bigcup but can't get the sum to work, even copying and pasting from the post above), so i'm just going to do a skeleton proof.

Let A1, A2,... be any infinite sequence of events, and let B1, B2,... be another sequence of events defined as follows: B1 = A1, B2 = Ac1A2, B3 = Ac1Ac2A3...
Claim: Pr(U from i=1 to n Ai) = Sum(from i=1 to n Pr(Bi)) for n = 1, 2,...
Proof: Base case: this is true for n = 1
Assume it is true for n = k-1.
So Pr(U from i=1 to k-1 Ai) = Sum(from i=1 to k-1 Pr(Bi)).
Then Sum(from i=1 to k-1 Pr(Bi)) + Pr(Bk = Sum(from i=1 to k Pr(Bi)) = Pr(U from i=1 to k-1 Ai) + Pr(Bk = Pr(U from i=1 to k-1 Ai) + Pr(Ac1Ac2...Ack-1Ak = Pr(U from i=1 to k-1 AiUAk) = Pr(U from i=1 to k Ai) = Sum(from i=1 to k Pr(Bi)).

So, by the principle of mathematical induction, the claim is proved.

is this more or less right? please excuse the notation, but i figured if i learned enough latex to do this post, it'd have taken me an extra hour (on top of that i already lost a proof when my computer didn't want to connect to the website and i had to retype it).

also, is there any reason you chose k-1 and k, and not k and k+1? just curious. thanks.

10. Jun 10, 2010

### lanedance

or how about (with no induction, sort of...)

pick n>m
$$B_n \cap B_m = (A_n \cap \bigcap_{i=1}^{n-1} A_i^c ) \cap (A_n \cap \bigcap_{j=1}^{m-1}A_j^c ) = 0$$

so the B's are independent, then
$$P(\bigcup_{j=1}^{n} B_j) = \sum_{j=1}^{n} P(B_j)$$

now the unions of the Bs is:
$$\bigcup_{j=1}^{n} B_j = \bigcup_{j=1}^{n} (A_j \cap \bigcap_{i=1}^{j-1} A_i^c) =\bigcup_{j=1}^{n}A_j$$