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Assume I have two random variables $A$ and $B$, which are not independent. In my particular case they will be values of a stochastic process at two given points in time, where $A$ is observed at an earlier time.

Define $(B|A)$ to be a conditional random variable, i.e. a random variable defined by the conditional distribution of $B$ given $A$. Question: is $(B|A)$ independent of $A$? Why? Why not? Under what conditions it is?

EXAMPLE: Let $A$ and $C$ be two independent Gaussian (0, 1) random variables, and let $B=A+C$. Then $(B|A=a)$ is Gaussian (a, 1) and seems to be independent of $A$, but I am not sure how to work formally with these kind of things.

EDIT: As Sebastian Andersson pointed out in the comment, $A$ and $(B|A)$ seem not to be independent. However, what if we condition on $A=a$, where $a$ is a constant? The intuition would be that first we are interested in the uncertain event $A$, and then after it happens (and we know the outcome), we are interested in an event $(B|A=a)$. Does it make sense?

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Define (B|A) to be a conditional random variable, i.e. a random variable defined by the conditional distribution of B given A... Such a thing does not exist, one can at most use the conditional distribution of B conditionally on A. Any source for your question? –  Did Apr 28 '13 at 12:58
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Interesting question. If you have that $X$ and $Y$ are independent, then $\int_Y f(x)f(y)dy=\int_Y f(x,y)dy=f(x)$. In your case, $\int_A f(b|a)f(a)da=\int_Af(a, b)da=f(b)\neq f(b|a)$. –  hejseb Apr 28 '13 at 13:00
    
@Did, I saw it here: maths.qmul.ac.uk/~pettit/MTH5122/notes15.pdf –  Grzenio Apr 28 '13 at 13:13
    
@SebastianAndersson, thanks for your comment! I added an extra paragraph to the question (not sure it makes sense though)... –  Grzenio Apr 28 '13 at 13:13
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If you look closely at this source you will see that they compute conditional distributions, never conditional random variables (although, unfortunately, they do use the term). My guess is that they (and you) would be in great trouble to actually define a notion of independence between A and (B|A) since there is no such thing as a random variable (B|A). Wouldn't you? –  Did Apr 28 '13 at 13:18
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As Did says, there is no (currently widely used) definition of a random variable $(B|A)$. I actually once spent some time thinking about whether such a thing could be defined, and came to the conclusion that it could, but that it wasn't a terribly useful definition. So let us say no more about that.

However, as you say, we can certainly consider the probability distribution of $B$ when conditioned on a particular value of $A$, which we could write $(B\mathop{|}A=a)$. (This notation isn't standard, but I don't think there's anything wrong with it, so I'll use it. Just remember that this notation stands for a probability distribution, not a "random variable" in the usual sense of the word.) In this case it's true that $(B\mathop{|}A=a)$ is independent of $A$. This is simply because the conditional probability $p(B=b\mathop{|}A=a) = p(B=b,A=a)p(A=a)$ assumes $A$ has the value $a$, regardless of its "actual" value. Thus if $A$ is measured and found to have the outcome $a'$, this does not change anything about $(B\mathop{|}A=a)$.

In the context of your stochastic process, the disribution $(B\mathop{|}A=a)$ has the interpretation "the probability distribution we would have at time $t_2$, if we observed the value $a$ at time $t_1$." You should be able to see intuitively that this doesn't depend on whether we've observed the system at time $t_1$, or what the result was if we did.

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