# Understanding $P(\bigcap_{i=1}^n A_i) = \prod_{i=1}^n P(A_i)$

1. Jul 31, 2012

### operationsres

My goal: Understand $P(\bigcap_{i=1}^n A_i) = \prod_{i=1}^n P(A_i)$.

My current understanding:

It wasn't defined in my lecture slides what $A_i$ exactly was, but I'm guessing that it's to be an event as that's how it's been defined everywhere else.

So, let A be an event (in this case, the outcome of 1 fair coin flip), and $\Omega$ be the state space.

We clearly have $\Omega = \{H,T\}$ and the $\sigma$-algebra as $\bf{F} = \{\{\},\Omega,\{H\},\{T\}\}$.

$\{H\},\{T\}$ are the only two events, so denote them $A_1 = \{H\}$ and $A_2 = \{T\}$.

We then have $\bigcap_{i=1}^2 A_i = \{ \}$ $\Rightarrow$ $\prod_{i=1}^2 P(A_i) = P(\{ \}) = 0$. Yet this is clearly incorrect as $\prod_{i=1}^2 P(A_i) = P(\{H\})*P(\{T\}) = 0.25$

_________

Now, I understand that this is to be applied to problems like ""You flip 2 fair coins, what's the probability that you get 2 heads"", then you multiply 0.5*0.5. But I'm really confused because $A_i$ was referred to throughout this whole lecture as constituting all the events that are elements of a relevant sigma-algebra.

What am I missing?

2. Jul 31, 2012

### uart

Hi operationres. That relationship only applies to independent events.

3. Jul 31, 2012

### uart

One way we can think about this is to rearrange the conditional probability formula (see your other recent post) as,
$$P(B \cap A) = P(A) P(B|A)$$.

Now if A and B are independent events then $P(B|A) = P(B)$, giving this result,

$$P(B \cap A) = P(A) P(B)$$.

4. Jul 31, 2012

### operationsres

Thanks, I completely understand now.

5. Jul 31, 2012

### Ray Vickson

The result stated applies only to independent events, and whether or not two events are independent depends on the nature of the probability measure P. In your coin example, we could define a probability measure P such that P{H} = 1, P{T} = 0, etc. Now we would have P{A & B} = P{A}*P{B} for all elements A and B of {{},Ω, {H}, {T}}. Of course, that probability measure does not look very useful; in practice, it would apply to a two-headed coin.

RGV