# Independence: Evaluating $\mathbb{P}(X_1 \in A_1, …, X_n \in A_n)$

In the context of independent random variables:

What is the meaning of $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$$

in

$$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)=\mathbb{P}(X_1 \in A_1)...\mathbb{P}(X_n \in A_n)$$?

Or, how does one evaluate:

$$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$$

without knowing the RHS.

Also,

Is

$$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$$

equivalent to

$$\mathbb{P}(X_1 \in A_1 \bigcap ..., \bigcap X_n \in A_n)$$

(since I've seen these both definitions for "independence of random variables")

(Note: I know how to evaluate the RHS of the equation, simply evaluate each probability independently, but not the LHS).

• The notation "$\mathbb{P}(X_1\in A_1, \ldots, X_n\in A_n)$" is a shorthand for the event $(X_1,\ldots,X_n)\in A_1\times \cdots \times A_n$: that is, each $X_i$ simultaneously lies in the corresponding $A_i$ for all $i=1,\ldots, n$. How the probability is actually "evaluated" depends on the form in which it is available to you, such as a probability function, a probability density, a (multivariate) characteristic function, or something else. In light of this, could you please edit the question to clarify what kind of answer it's looking for? – whuber Nov 7 '15 at 14:44

The generic meaning of $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$$ is the measure of the $n$-dimensional set $A_1\times A_2\times\cdots\times A_n$ under the probability measure $F_n$ associated with the joint distribution of the random vector $(X_1,\ldots,X_n)$. That is, $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)= \int_{A_1\times\cdots\times A_n}\text{d}F_n(x_1,\ldots,x_n)\,.$$ In the event this joint distribution has a density $f_n$ wrt the Lebesgue measure on $\mathbb{R}^n$, it also writes as $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)= \int_{A_1\times\cdots\times A_n}f_n(x_1,\ldots,x_n)\text{d}x_1\cdots \text{d}x_n\,.$$ In your special case [that later got removed from the question!], things simplify drastically: $$\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)=\mathbb{P}(\omega \in A_1, ..., \omega \in A_n)=\mathbb{P}(\omega \in A_1\cap\cdots\cap A_n)$$which is equal to$$\dfrac{\text{card}(A_1\cap\cdots\cap A_n)}{n}$$
The equation states that in case the variables $X_{1}, ..., X_{n}$ are all independent, then the joint probability: $P(X_{1}, ..., X_{n})$ is given by the product of the single probabilities: $P(X_{1})P(X_{2})...P(X_{n})$ which can be derived by applying the chain rule to the join probability.
If the variables $X_{1}, ..., X_{n}$ would not be independent then when applying the chain rule you would get that $P(X_{1}, ..., X_{n})$ = $P(X_{1}, ..., X_{n-1} \| X_{n})P(X_{n})$ which you can then recursively apply to all joint terms.
• But how does one display that $\mathbb{P}(X_1 \in A_1)...\mathbb{P}(X_n \in A_n)=\mathbb{P}(X_1 \in A_1, ..., X_n \in A_n)$ (starting from the LHS), when one does not yet know whether $X_1,...,X_n$ are independent. – mavavilj Nov 7 '15 at 14:37
• Theoretically you can't do that. That is why you always make assumptions about your variables. Empirically however (when you have data available) you can of course compare $P(X_{1})$ to $P(X_{1}\|P(X_{2}))$ and see if they differ over your data set. – Sjoerd Nov 7 '15 at 14:41
• Such as? Is it an assumption to claim that each $X_i$ follows a certain probability distribution? – mavavilj Nov 7 '15 at 14:43
• In case of real data you can empirically test it. In case of theory it would depend on your problem (i.e. the characteristics of $X$) as to whether it is relevant to make an assumption that they are independent. If X for example follows a Bernoulli distribution, then that would yield independence. – Sjoerd Nov 7 '15 at 14:45