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19h
comment Conditional probability with sigma-field
And what are your thoughts on this exercise?
Feb
2
answered Expectation of min(X,1)
Feb
1
comment How to understand $E(X\mid B)$ in the measure theory way
@Did: Sorry, for $P_X$-almost-all $x$.
Jan
29
comment How to understand $E(X\mid B)$ in the measure theory way
That's defined by Did in a previous comment. And you're right, it should be $x$.
Jan
28
comment How to understand $E(X\mid B)$ in the measure theory way
@JohnZHANG: If $P(B)>0$ then $E(Y\mid B)=E(Y\mathbf{1}_B)/P(B)$.
Jan
19
comment If $E(X^-)=0$ then $P(X\geqslant0)=1$
And I'm giving you hints to try and involve you in the process thinking you might learn something.
Jan
19
comment If $E(X^-)=0$ then $P(X\geqslant0)=1$
This gives you that $I^{-}(t)=0$ almost surely and hence...
Jan
19
comment If $E(X^-)=0$ then $P(X\geqslant0)=1$
In general, if $X$ is non-negative and $\mathrm{E}[X]=0$, then $X=0$ almost surely.
Jan
19
comment Calculate $\text{Var} [-3Z ] $
Looks good to me.
Jan
16
comment Convergence in probability implies weakly convergence of sequence of induced measures
@User3101: Then why did you ask the question?
Jan
15
answered Convergence in probability implies weakly convergence of sequence of induced measures
Jan
14
comment Empirical Distribution converges to the distribution in $L^1$?
Hint: Show that $(F_n(x))_{n\geqslant 1}$ is uniformly integrable.
Jan
6
comment Probability integration notation: integrating with respect to X over a set involving $Y$
OP didn't mention the existence of a joint density.
Jan
6
comment Probability integration notation: integrating with respect to X over a set involving $Y$
Should have been $(Y\in B)$ instead of $(Y\in A)$.
Jan
6
comment How to prove Boole’s inequality
@mavavilj: In general, you can only say that $P(B_n)\leq P(A_n)$ when $B_n\subseteq A_n$. Whether it is an equality or strict inequality would depend on the actual $A_n$ and $B_n$. However, we don't really need to this information to conclude the proof.
Jan
6
comment How to prove Boole’s inequality
@mavavilj: It holds since $B_n\subseteq A_n$ for all $n$.
Jan
5
comment How can I explain $0.999\ldots=1$?
@user170039: Sure, you could also define to be equal to $3$. In my post I define to be the limit of the sequence $(q_n)$ or alternatively the number being $\leqslant 1$ and $\geqslant q_n$ for all $n$, and hence implicitly assuming that it's a real number. But so do you when you define to be equal to $1$. So what exactly is your point?
Jan
5
comment How can I explain $0.999\ldots=1$?
@user170039: If it's not given that it should be a real number, then you tell me how one should define or characterize it. That would be like saying 'Here is a quantity I'll call $x$, but I'm not willing to say anything about it. Now you tell me what it is'.
Jan
5
comment How can I explain $0.999\ldots=1$?
@user170039: I defined $0.999\ldots$ as the limit of the sequence $(q_n)$ as is written in the beginning of my post. Or you could say that $0.999\ldots$ is the number satisfying $0.999\ldots\leqslant 1$ and $0.999\ldots\geqslant q_n$ for all $n$ from which it follows that it's $1$.
Jan
4
reviewed Looks OK $4^\text{th}$ order polynomial approximation for: $f(x) = e^{-\cos(2\pi x)}$