Reputation
17,462
Top tag
Next privilege 20,000 Rep.
Access 'trusted user' tools
Badges
7 21 45
Newest
 Nice Answer
Impact
~207k people reached

Jul
22
awarded  Nice Answer
Jul
20
comment Expectation with exponential random variable
@Chival: $\mathrm{E}[Y\mid A]=\mathrm{E}[Y\mathbf{1}_A]/P(A)$ for any event $A$ with $P(A)>0$. Hence you have an indicator function too much.
Jul
20
comment Expectation with exponential random variable
@Math1000: It is a number, since we're conditioning on a set, not a sigma-algebra.
Jul
17
answered Deriving density of a function of a random variable
Jul
17
answered Splitting the summation sign
Jul
17
answered Why Does $\rm{E}[1_A \mid X] = P^X(A\mid X)$?
Jul
16
comment What is meant by $P(X = x, Y = y)$?
$P_X$ is by definition given by $P_X(A)=P(X^{-1}(A))$ for $A$ being a Borel set and hence $P_X(X^{-1}(x))$ doesn't make sense. $P_X$ is a measure on $(\mathbb{R},\mathcal{B}(\mathbb{R}))$ while $X^{-1}(x)$ is a subset of $\Omega$ and hence cannot be measured by $P_X$.
Jul
15
comment On the central limit theorem
You're right, it doesn't make sense since the righthand side depends on $n$.
Jul
15
comment Plugging a random variable into a probability function?
What definition of $\mathbb{E}(X\mid Z)$ are you working with?
Jul
15
revised What's the name of this theorem?
added 19 characters in body
Jul
15
revised What's the name of this theorem?
added 51 characters in body
Jul
15
comment What's the name of this theorem?
@Ant: You need some kind of topological structure in order to talk about open sets and continuity. I have never heard this theorem go under any name by itself, but it's a consequence of the uniqueness of densities which again is a consequence of the uniqueness theorem of measures.
Jul
15
revised What's the name of this theorem?
added 22 characters in body
Jul
15
answered What's the name of this theorem?
Jul
14
comment Simple Set Operation with Random Variable
The left-hand side is either $\Omega$ or $\varnothing$ depending on whether $1-a\leq \varepsilon$ or not.
Jul
6
awarded  Nice Answer
Jun
3
comment Why is this wrong - conditional expectation of brownian motion: $\mathbb{E}[B_1 | B_2]$
@elbarto: The Brownian motion has independent increments, in particular, $B_2-B_1$ is independent of $B_1$ and hence $\mathbb{E}[B_2-B_1\mid B_1]=\mathbb{E}[B_2-B_1]$.
May
28
comment Expectation of $\mathbb{E}(X^{k+1})$
@Michael: Sure, I just didn't agree with what you were saying. My comments weren't meant in any harsh way :)
May
28
comment Expectation of $\mathbb{E}(X^{k+1})$
@Michael: People usually mean Tonelli when the function is non-negative, so there's no need for further justification here.
May
28
comment Expectation of $\mathbb{E}(X^{k+1})$
@Michael: For one, you would have to show that the limit in your expression is in fact 0. I think the measure-theoretic approach is mathematically much neater and essentially the exercise is just a simple application of Fubini.