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6h
comment Computing the limit of $(\lg n)^{0.5}/\lg(n^{0.5})$
THAT would be a fruitful idea, yes...
6h
comment Proving a friend wrong about continuity
"I figure the solution is to prove the negation. i.e. prove..." Good idea. What is the negation, already (you copied the assertion itself, it seems)?
6h
comment On a proof regarding the sigma algebra generated by a single random variable.
Simply read the definition 5 pages earlier, at the beginning of section 3.8...
6h
comment On a proof regarding the sigma algebra generated by a single random variable.
Indeed, 5 pages earlier, at the beginning of section 3.8...
8h
comment Show that $ H_n \leq H_{\left\lfloor\frac{n}{2}\right\rfloor}+1$
$|A_\color{red}{3}|=2>\frac{\color{red}{3}}2$.
8h
comment PDF of $Y=\min(0,X)$ when PDF of $X$ is $\frac34(1-x^2)$ on $(-1,1)$
Dunno how you arrived at this result but indeed X=2Z-1 where Z is Beta(2,2), with density 6z(1-z) on (0,1).
9h
comment Conditional distribution of mixed process
OP: The interpretation of your model in the answer below is quite different from the one in my comment. No problem about that (since here we are all guessing what you mean) but you might want to explain clearly as soon as possible which interpretation is correct. Note that you state in your question that the process does not have independent increments while the model suggested in the accepted answer has.
10h
comment Conditional distribution of mixed process
Since this question seems in dire need of precision, here is a construction of $N$ (if you have something different in mind, at least we will have something solid to discuss): consider some independent Poisson processes $N_0$ with rate $\lambda_0$ and $N_1$ with rate $\lambda_1$ and Bernoulli random variable $B$ such that $P(B=1)=p=1-P(B=0)$, then $N(t)=BN_1(t)+(1-B)N_0(t)$ for every $t$. Thus, for $s<t$, $P(N(s)=k)=pP(N_1(s)=k)+(1-p)P(N_0(s)=k)$ and $P(N(s)=k,N(t)=\ell)=pP(N_1(s)=k,N_1(t)=\ell)+(1-p)P(N_0(s)=k,N_0(t)=\ell)$. From there, simply proceed with the computations.
10h
comment Uniform PDF for continuous variable, why does the probability values increase to 1, when its normalized?
Why $ Pr[X = x] = \frac{1}{2\pi}$?
10h
comment PDF of $Y=\min(0,X)$ when PDF of $X$ is $\frac34(1-x^2)$ on $(-1,1)$
Hint: When is $Y=\min(0,X)$ equal to $X$? When is it equal to $0$?
12h
comment Death process - median time to die out
Sorry but (W_n <= t) is not (X(t) >= n) but (X(t) <= X(0)-n).
13h
comment Help needed to solve probability problem
?? No this is not P(h)P(t)P(t), yes this is 3h(1-h)^2. (And please use @.)
13h
comment Conditional Expectation
As explained below, this assumes implicitely that $(X,Y):(x,y)\mapsto(x,y)$ on $\Omega=[0,1]^2$. What is your source?
14h
comment Show that the function is continuous
"To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ ... is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$" Absolutely not true (even replacing $(x_0,y_0)$ by $(0,0)$).
14h
comment Adding two discrete distributions
@AndrewB. That you chose to accept the answer you chose to accept seems to indicate you do not understand the exercise. One might regret it.
14h
comment Adding two discrete distributions
Not Z = X but distribution of Z = distribution of X.
14h
comment Four Conditional Expectation Problems
Thanks for the reference (not very usable, though). // The trouble with your approach of the site is that you dump lists of exercises here and ask for their complete solution without any personal input. Did you try to read some howtoask page? (And please start using @ now.)
14h
comment On the continuity of Li's numbers.
Rather stunning discussion above, I must admit... @Tatenda: Here is a simple question for you: assume hypothetically for a moment that $\lambda_n=2^n$ for every $n$, do you call this "a continuous function of $n$"? Note that since the objections made to you about the undefined notion of continuity of a sequence you seem to want to promote are quite basic, failing to address them (as you did, so far) can only convince everybody that you are either a crank or a troll. I know, life is hard.
16h
comment Does there exist any unbounded above function $f(x)$ such that $f(x)<\log(x)$ for all $x>M$
And why this exercise should prevent you from accepting an answer (if this is indeed your position) is beyond me, I must admit.
16h
comment Does there exist any unbounded above function $f(x)$ such that $f(x)<\log(x)$ for all $x>M$
@DEEP Think. Deeply.