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Apr
15
comment How do I compute $P(X=Y)$? for independent random variables with with geometric distribution.
Hint: $[X=Y]=\bigcup\limits_{k=0}^\infty [X=k,Y=k]$
Apr
14
comment Let $U=2X$ and $V=-3Y$. Find Correlation (U,V) given Correlation $(X,Y)=0.8$.
Looks fine to me.
Apr
5
reviewed Looks OK Can modulo(remainder) be distribute over division?
Apr
5
reviewed Looks OK Is there a notation for a set of angles?
Apr
3
comment Indicator function and expectation
So $x$ is both uniformly distributed and $x$ is something you have drawn, i.e. a constant? Please, make your question more clear.
Apr
3
comment Indicator function and expectation
$\mathbb{P}(1_{x\geq \tilde{x}})$ doesn't make sense since $\mathbb{P}$ assigns a number to a set not a function.
Mar
30
comment Factor of a joint density into product of functions
If both $g$ and $h$ are non-negative, then $f_X(x)=(\int_\mathbb{R} g(x)\,\mathrm dx)^{-1} g(x)$.
Mar
26
comment Finding the conditional distribution of a normal RV given another normal RV
$(Z,W,V)$ is jointly normal by the independence assumption and $(X,Y)$ is just a linear transformation of $(Z,W,V)$ and hence is also jointly normal.
Mar
26
comment Measure-theoretic conditional expectation
It's the other way around that holds: if $f$ is $\mathcal{N}$-measurable and $\mathcal{N}\subseteq\mathcal{M}$, then $f$ is also $\mathcal{M}$-measurable.
Mar
26
comment Showing this random variable has normal distribution
By definition, a random variable $X$ follows a (standard) normal distribution if $P(X\in A)=\int_A\frac{1}{\sqrt{2\pi}}\mathrm{e}^{-x^2/2}\,\mathrm dx$ for all Borel sets $A$. Now you just have to argue that $P(X\in A)=P(A)$ which should be fairly obvious.
Mar
25
comment A doubt regarding integration of simple Random variables
Just divide the $A_i$'s further. There is no requirement that all $a_i$'s should be different. In your example, split $A_1$ into the two disjoint sets $\tilde{A}_1$ and $\tilde{A}_2$ in which $Y$ takes on different values. Then you would have $b_1=5$ and $b_2=10$ for $Y$ and $a_1=a_2$ for $X$.
Mar
24
comment $P(B)$ given $P(A\cup B)$ and $P(A'|B')$: Probability
@Kayseta: Exactly.
Mar
24
answered $P(B)$ given $P(A\cup B)$ and $P(A'|B')$: Probability
Mar
24
comment Probability distribution of $Z=X_1\,I(U<p) + X_2\, I(U\ge p)$?
By definition, $Z=X_1$ on the set $[U<p]$ which has probability $p$ of occuring and $Z=X_2$ on the set $[U\geq p]$ which has probability $(1-p)$ of occuring.
Mar
24
comment Probability distribution of $Z=X_1\,I(U<p) + X_2\, I(U\ge p)$?
Yes, the distribution function of $Z$ is $pF_1+(1-p)F_2$ regardless of whether $X_1$ and $X_2$ are independent or not.
Mar
24
comment If $P(A) = 0$, then can we conclude that $Q(A) = 0 $ ??
@Jack: Yes, that's an easy consequence of the monotone convergence theorem.
Mar
24
comment If $P(A) = 0$, then can we conclude that $Q(A) = 0 $ ??
Yes, real numbers are random variables and of course $Q(A)=0$ also holds almost surely (with respect to any probability measure) since it actually holds everywhere, but I doubt this is what the OP meant. The point I was trying to make was that constants are either equal to 0 (everywhere) or not and hence the statement "$Q(A)=0$ almost surely" suggests that the OP has misunderstood something. I say with respect to $P$ to emphasize which probability measure the statement "almost surely" holds for. Yes, $Q$ is a probability since it's assumed that $\mathbb{E}_P[X]=1$
Mar
24
comment If $P(A) = 0$, then can we conclude that $Q(A) = 0 $ ??
No it doesn't. $Q(A)$ is not random, it's a real number.
Mar
24
comment If $P(A) = 0$, then can we conclude that $Q(A) = 0 $ ??
Because $\mathbf{1}_A=0$ almost surely (with respect to $P$).
Mar
24
answered If $P(A) = 0$, then can we conclude that $Q(A) = 0 $ ??