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16h
comment How to find the density of $Y=g(X)$ in this case?
You can't. The three random variables in $Y$ are not jointly continuous and as such do not enjoy a joint density function $f_{Y_1,Y_2,Y_3}(y_1,y_1,y_3)$. On a smaller scale, note that $W$ and $Z=W^2$ are two real random variables that do not possess a joint density because all the probability mass lies on the curve $z=w^2$ in the $w$-$z$ plane and the density (measured in mass per unit area) of the probability mass at points on this curve is unbounded.
1d
comment Proof that Sum of $n$ Squared Errors ~ Chi Square with $n$ $df$
One of the forms of the Fundamental Theorem of Calculus deals with the derivative of a function whose value is the value of an integral, and the function argument appears in the limits, If $$F(\alpha)=\int_{a(\alpha)}^{b(\alpha)}f(x)\mathrm dx$$, then $$\frac{\mathrm dF(\alpha)}{\mathrm d\alpha}=f(b(\alpha))\frac{\mathrm db(\alpha)}{\mathrm d\alpha}-f(a(\alpha))\frac{\mathrm db(\alpha)}{\mathrm d\alpha}.$$ So, apply this result.
1d
comment sufficient conditions for a stochastic process to be wide sense stationary
You should include what you think are the definitions of the autocorrelation function and autocovariance function of a random process. Your "familiar form" $$R(\tau)=\frac{E[(X_t-\mu)(X_{t+\tau}-\mu)]}{\sigma^2}$$ would be called the normalized autocovariance function by those who define the autocorrelation function as $R(t_1, t_2) = E[X_{t_1}X_{t_2}$ and for wide-sense-stationary processes, $$R_X(\tau) = R(t,t+\tau) = R(0,\tau).$$
2d
comment Sufficient condition for $E(wu\mid v)=0$ given that $E(u\mid v)=0$?
The conditional covariance of $W$ and $U$ given that the random variable $V$ has value $v$ is \begin{align} \operatorname{cov}(W, U\mid V = v) &= E[WU\mid V=v] - E[W\mid V=v]E[U \mid V = v]\\ &= E[WU\mid V=v] - E[W\mid V=v]\times 0\\ &= E[WU\mid V=v]. \end{align} Thus, independence might not be needed; it would suffice if $W$ and $U$ were conditionally uncorrelated random variables given $V$.
2d
answered Understanding step in derivation of joint distribution
Aug
2
comment How and why do rumors/gossip spread?
People have looked at how gossip spreads, though perhaps not with the specific properties that you wish to include.. See for example, A.G. Dimakis, A.D. Sarwate, M.J. Wainwright, Geographic Gossip: Efficient Averaging for Sensor Networks, IEEE Transactions on Signal Processing 56(3): pp. 1205--1215, March 2008, or T.C. Aysal, M.E. Yildiz, A.D. Sarwate, A. Scaglione, Broadcast Gossip Algorithms for Consensus, IEEE Transactions on Signal Processing 57(7): pp. 2748--2761, July 2009. and no, that A.D.Sarwate is not me.
Jul
30
comment Can an interval be represented as a set?
Is $a,a,a,a,\cdots$ a sequence whose limit is $a$, and if so, do we exclude this sequence from consideration when we say "A sequence is said to converge to a limit $a$ if $\ldots$" ? Nothing in that last example contradicts set continuity.
Jul
29
comment Questions about Variance and Covariance
var$(aX+b) = a^2$var$(X)$ is a result that you should commit to memory.
Jul
28
comment Can an interval be represented as a set?
Much of elementary calculus can be said to be a study of continuous functions: those that enjoy the property that for each and every sequence of numbers $\{x_i\}$ converging to a limit $a$, the sequence $\{f(x_i)\}$ converges to $f(a)$, the value of the function $f$ at $a$, i.e. $f($limit$)$ equals limit$(f)$. The probability measure $P(\cdot)$ maps sets to real numbers. For every sequence of sets $\{A_i\}$ that approach a limit $A$, the sequence $P\{A_i\}$ also approaches a limit, and that limit is $P(A)$. Thus, $P(\cdot)$ is said to be set-continuous: $P($limit$)$ equals limit $(P)$
Jul
28
comment Can an interval be represented as a set?
@user50224 Yes, I am still right.
Jul
28
comment Can an interval be represented as a set?
An interval is a set: a set of real numbers. The "previously proved result" $$P\left(\lim_{n\to\infty} \bigcup_{i=1}^n A_n\right) = \lim_{n\to\infty}P\left(\bigcup_{i=1}^n A_n\right) $$ is actually equivalent to the third axiom of probability. I like to think of it as saying that probability is a set-continuous function: a real function $f(x)$ is continuous at $a$ if for every sequence $x_1,x_2,\ldots, x_n, \ldots$ that converges to $a$, the sequence $f(x_1), f(x_2),\ldots, f(x_n),\ldots$ converged to $f(a)$.
Jul
27
comment How to understand $E(XY)$ intuitively
@Lzy See the penultimate paragraph in the answer modified as below: Yes, $Z=h(X,Y)$ is a random variable $Z$ in its own right and the definition of $E[Z]$ is just $(3)$, but gLOTUS allows us to bypass the step of pre-computing $f_Z(z) $but use instead the double integral $$\int_{-\infty}^\infty \int_{-\infty}^\infty h(x,y) f_{X,Y}(x,y) \,\mathrm dy\, \mathrm dx= \int_{-\infty}^\infty \int_{-\infty}^\infty h(x,y)f_{X,Y}(x,y) \,\mathrm dx\, \mathrm dy$$
Jul
26
answered How to understand $E(XY)$ intuitively
Jul
24
comment Can I compute marginal distribution this way?
Since $F_{X\mid Y}(x\mid y) \leq 1$, your formula $$F_X(x)=\int^{F_{x\mid y}(x\mid y)}_{-\infty}f_Y(y)\,dy$$ looks highly suspect because clearly $F_X(x)$ has maximum value $$\int_{-\infty}^1 f_Y(y)\,dy = F_Y(1).$$
Jul
24
revised Independence, conditioning, and correlation part 2
added 572 characters in body
Jul
24
comment Can I compute marginal distribution this way?
Which book are you reading?
Jul
24
answered Sum of two truncated normaly distributed variables
Jul
24
answered Independence, conditioning, and correlation part 2
Jul
23
comment Equivalence of Conditional Expectations w.r.t. Discrete Random Variable
$P(Y=y) >0$ for all $y \in \mathbb R$? Do you really mean this?
Jul
23
comment What is the intuition behind the probability of the first even toss of a die tossed repeatedly and independently?
Partition the successive rolls of the die into sets of 2 successive rolls. As long as a set of 2 successive rolls results in (O,O), you will roll the die again. Since what is past is prologue, let is look at the very first set of two rolls on which at least one E occurs. There are three equally likely possibilities are (E,O), (E,E), and (O,E), and only in the third case (O,E) does the first E occur on an even-numbered roll. So, the desired probability is 1/3.