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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.
May
28
comment Expectation of $\mathbb{E}(X^{k+1})$
The right-hand side doesn't make sense. First of all, there are three integrals there (I suspect the innermost integral should have been an indicator function). Second, the outermost integral is over $[0,\infty)$ but you're integrating with respect to $P$ which is a measure on $(\Omega,\mathcal{F})$. @Michael: That is essentially making things harder even at the cost of generality.
May
20
comment $\sigma$-algebra
If $U\subset A$, then $U=A\cap U$ which is in $A\cap\Sigma$.
May
19
comment $\sigma$-algebra
That $A\cap \Sigma\subset \Sigma$ means simply that for every $B\in A\cap \Sigma$, we also have that $B\in \Sigma$. So this makes sense to write.
May
19
comment $\sigma$-algebra
Since $\Sigma$ is a $\sigma$-algebra, then $A\cap M\in\Sigma$ for all $M\in\Sigma$, no?
May
17
revised What is the name of this theorem, and are there any caveats?
added 195 characters in body
May
17
comment What is the name of this theorem, and are there any caveats?
Yeah, you're right @drhab. Thanks for pointing that out.
May
13
comment How to calculate a population mean for a normal distribution
Yeah, you're right. Brainfart by me.
May
13
comment How to calculate a population mean for a normal distribution
It's the general formula for the expectation of a discrete random variable. No, the variables doesn't follow a normal distribution since they are discrete and the normal distribution is a continuous distribution. In fact, they follow a discrete uniform distribution assigning equal probability to each point in their support. The CLT may however be used to approximate the distribution of $\bar{X}$ (which is also discrete) for large $n$ by a normal distribution.
May
13
comment How to calculate a population mean for a normal distribution
But $X_1,\ldots,X_{50}$ doesn't follow a normal distribution. You've specified their distribution earlier as $P(X_i=x)=x/55$.
May
12
comment Expected value of a function of a random variable
"P(dω) is f(x)dx" - It's not. "f(x)=α|α∈(−∞,x)∧α=P[ω:X(ω)≤x]" - What? "The distribution of Z is αg−1α" - How does $\alpha g^{-1} \alpha$ even make sense? It's also a big thing just to take the "change of variable" theorem for granted.
May
8
comment $X + Y \overset{\mathcal{D}}{=} X \Longrightarrow \mathbf{P}[Y = 0] =1$
You can't be sure that $\phi_X(t)\neq 0$ for all $t$ and therefore can't divide by it for all $t$. However, we know that $\phi_X(t)\neq 0$ for $t\in (-\varepsilon,\varepsilon)$ for some $\varepsilon>0$ (why?) and hence $\phi_Y(t)=1$ for $t\in (-\varepsilon,\varepsilon)$. This is actually enough to conclude that $Y$ is constant almost surely.
May
7
comment Finding all of the solutions to the trigonometric equation $\cos (t \Omega )=\frac{x}{a}$
Cosine is a periodic function and hence if your equation has a solution, then it has in fact infinitely many solutions.
May
7
comment Difference between conditional expectation and conditional probabilty
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 $P(A\mid B)=P(A\cap B)/P(B)$?
May
6
comment Formal definition of conditional probability
@simonzack: Take a look at this.
Apr
24
comment Find distribution of rv X_N where N is independent rv and each X_i~exp(\lambda_i)
Please take a look at the MathJax formating now, so you are able to properly format your question next time. Also, the question probably reads: "Let $X_1,X_2,\ldots$ be..."