These are known definitions: We have a probability space $(\Omega, A, P)$

Conditional probability is defined through $P(A|B) = \frac{P(A \cap B)}{P(B)}, P(B) > 0$. This is a real nunmber.

Then also where is conditional expectation $E[X|A_0]$ with $A_0$ being some subalgebra. This is a random variable. In the special case when $A_0$ is generated by a random variable $Y$ we also write $E[X|\sigma(Y)] = E[X|Y]$ and using the factorization theorem you can also write $E[X|Y]$ as a RV of $Y$ which you then denote as $E[X|Y=y]$. For $X$ being in indicator function we sometimes also write $E[1_A | Y=y] =: P(A | Y = y)$ which confuses me very much, because this now is the evaluation of an only a.s. defined RV

I am particular confused with the proof I state in this Question.

Since the probability of $\{X_1 = i_1, ... X_{n-1} = i_{n-1}\}$ is 0 because the $X_i$ are real valued, the proof can't be talking about conditional probabilities. Thus I think the proof uses the sloppy notation for conditional expectation $E[1_B(X_n) | X_1 = i_1, ... X_{n-1} = i_{n-1}] =: P(X_n \in B | X_1 = i_1, ... X_{n-1} = i_{n-1})$. Now in this concept, I dont understand why the single steps in the proof are true. They are all obvious true for $P(|)$ being conditional probability, but in my eye not trivial for $P(|)$ being conditional expectations

  • 1
    $\begingroup$ "and because we can factorize the conditional expectation through Y, we now can write E[X|Y=y]" Huh? Your trouble here and in your previous question might be due to an unsufficient distinction between the random variable $E(X|Y)$ and the real valued function $y\mapsto E(X|Y=y)$. Please realize that the basic object here (and the only one always well defined) is the random variable $E(X|Y)$. $\endgroup$ – Did May 7 '15 at 7:32
  • $\begingroup$ @Did, I clearified on that and also tried to make the question a bit more clear to what I need to know and what I don't understand $\endgroup$ – fubal May 16 '15 at 15:42
  • $\begingroup$ Am I just so much on the wrong track or does nobody know? I have a feeling I got the rest of the lecture by now, but this one proof doesn't get to me $\endgroup$ – fubal May 29 '15 at 12:09

Let $X, Y$ be random variable, then the condition expectation is given by $$ E(Y|X=x)=\int_{Y}yf_{Y|X=x}(x,y)dy=\int_{Y}y\frac{f_{X,Y}(x,y)}{f_{X}(x)}dy=\int_{y\in Y}y\frac{f_{X,Y}(x,y)}{\int_{t\in Y}f_{X,Y}(x,t)dt}dy $$ where $x$ is any element in $X$. Therefore $E(Y|X=x)$ is function of $x$. Here the inner term $f_{Y|X}(x,y)$ is precisely coming from conditional distribution $P(A|B)=\frac{P(A\cap B)}{P(B)}$. One way to think about it is to consider the discrete case, then we have $P(Y=y|X=x)$ given by the quotient $\frac{P(Y=y, X=x)}{P(X=x)}$. And $P(X=x)=\sum_{y\in Y}P(X=x, Y=y)$.

For what you asked on the second part, I think it is important to note that for Markov chains, the probability transition function from one state to the next state is independent of the previous states. You can think about this as tossing a fat coin on the real line. Suppose the coin goes to position $x$ after $n$-steps, whether the coin goes to $x\pm 1$ in the next step would be totally independent of its past. In other words, given the coin has undergone $HTHTTT\cdots $sequence does not change the expectation we have given the $n$-th step it is at position $x$ at all. The coin does not remember that. To me this is what makes the Markov chain so nice and useful.

  • $\begingroup$ Conditional expectation exists more generally than just the discrete or continuous case. And what do you mean by $f_{Y\mid X}(x,y)$ is precisely coming from $P(A\mid B)=P(A\cap B)/P(B)$? $\endgroup$ – Stefan Hansen May 7 '15 at 7:28
  • $\begingroup$ It is rather disquieting to see (as in the first line of this post) computations of conditional expectations reduced to the only case when a joint PDF exists. No, typical random variables on R^2 do not have a PDF and no, this is not how to correctly define E(X|Y). That this is routinely done on the site makes one suspect that this only reflects the flawed approach to the subject these users were subjected to. $\endgroup$ – Did May 7 '15 at 7:29
  • $\begingroup$ @Did: I totally agree. This sloppy approach is not what I really wanted, but being taught when I took the class myself. $\endgroup$ – Bombyx mori May 7 '15 at 15:23

It is true that if $X$ is an indicator of $A$, then the expected value happens of $X$ is the probability of $Z \in A$ (and this is true regardless of whether one conditions on $Y$). But that is not equivalent to saying that the definition of (conditional) expected value is equal to the definition of (conditional) probability. You have simply demonstrated equality under a specific situation; in other words, you have applied the definitions to that situation.

  • $\begingroup$ How can the expected value of $X$ conditioned on $Y$ be the be the probability of $Z \in A$, if the first is an a.s. defined random variable and the second is a real number between 0 and 1? $\endgroup$ – fubal May 16 '15 at 15:23

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