-1
$\begingroup$

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).

$\endgroup$
  • 1
    $\begingroup$ 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? $\endgroup$ – whuber Nov 7 '15 at 14:44
2
$\begingroup$

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$$\dfrac{\text{card}(A_1\cap\cdots\cap A_n)}{n}$$

$\endgroup$
1
$\begingroup$

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.

https://en.wikipedia.org/wiki/Joint_probability_distribution

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – mavavilj Nov 7 '15 at 14:37
  • 1
    $\begingroup$ 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. $\endgroup$ – Sjoerd Nov 7 '15 at 14:41
  • $\begingroup$ Such as? Is it an assumption to claim that each $X_i$ follows a certain probability distribution? $\endgroup$ – mavavilj Nov 7 '15 at 14:43
  • 1
    $\begingroup$ 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. $\endgroup$ – Sjoerd Nov 7 '15 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.