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Jun
29
comment Solve this logarithmic equation: $2^{2-\ln x}+2^{2+\ln x}=8$
The first method was modified. Now I fail to understand why $2^{\ln\:x}\:(2-2^{\ln x})=2^0\cdot 1$ implies $x=1$ (last step). Since this is the heart of the matter, I consider this first method as not being a proof. The second method is allright.
Jun
29
comment On the equality $p(A) = \int_{x} P(A|X=x)\ dF(x)$ in probability
If $p(x)=P(A\mid X=x)$ for every $x$ then indeed $p(X)=P(A\mid X)$. Thus, each $p(x)$ is a number and $p(X)$ is a random variable.
Jun
29
comment Show that the sequence $\left\{\frac{2n}{2n-1}\right\}$ is monotone by using $a_{n+1} - a_{n}$
Or that the sequence itself is?
Jun
29
comment On the equality $p(A) = \int_{x} P(A|X=x)\ dF(x)$ in probability
No, yours does not work because the third equality does not "follow from the definition of $p(x)$" (neither the fourth, since $p(X)$ is quite non deterministic).
Jun
29
comment process stochastics and branching process
Deliberate duplicate. Do not do that.
Jun
29
comment Is a martingale with bounded variance therefore bounded in $L^2$?
Hint: Can the sequence E(W_n) be unbounded?
Jun
29
revised Can we use Rolle's Theorem if $f : [a, b] \to \mathbb{R}$ is continuous on $(a, b)$ only?
rolled back to a previous revision
Jun
29
comment $E(f(|X_n|))$ property implies uniform integrability?
So that you can show something: how to solve the exercise if $f(x)=x^2$? Then if $f(x)=x\log x$?
Jun
29
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
As I said: CLT in the title, CLT in the body, CLT in your comment. But I am not sure that you are reading what I write anymore... Sorry for the disturbance.
Jun
29
comment Non linear Differential Equation
A good reason to use $\dot z/z^2$, and more generally $A/B$ rather than $\frac{A}B$, when not in displayed mode.
Jun
29
comment Probability that after 10,000 steps (+-1) you'll end up at the origin. How to use Central Limit Theorem?
What should you do? At least indicate that this does not follow from CLT, contrarily to what the OP asks in the title and in the body of the question and to what you unfortunately assert in a comment above, but from other arguments. To sum up, neither make people ignore this (no risk...) nor make them believe the justification is simply the CLT.
Jun
29
comment A prob =a random variable
I suspect this is offtopic, see comment.
Jun
29
comment Why is this the solution?
Even simpler: $\frac12\mathbb N$ is countable hence $P(X\in\frac12\mathbb N)=0$, and $[2X=3Y+1]\subseteq[X\in\frac12\mathbb N]$.
Jun
29
comment proving that $\lim_{n\to \infty}P(A_n)$ exists and $\lim_{n\to \infty}P(A_n) =P(\lim\sup A_n)$
As was already mentioned to this OP, their statement of the "second" Borel-Cantelli lemma is incorrect.
Jun
29
revised proving that $\lim_{n\to \infty}P(A_n)$ exists and $\lim_{n\to \infty}P(A_n) =P(\lim\sup A_n)$
deleted 144 characters in body
Jun
29
comment Conditional probability branching process
No need to find a book where they do that, just do it.
Jun
29
comment A prob =a random variable
@ncmathsadist More caution, please.
Jun
29
comment A prob =a random variable
Almost sure that $P(A)=E(I_A)$ is not the question. To prove is the more interesting implication that if $P(A)=I_A$ almost surely then $P(A)=0$ or $P(A)=1$.
Jun
29
comment How to prove that $f(x)$ is constant if $|f(x)-f(y)|^2 \le (x-y)^3$?
Interesting variation on the more usual (and asked several times on the site) condition that $|f(x)-f(y)|\leqslant C\cdot(x-y)^2$. The take-home message is that no differentiation is needed.
Jun
29
comment proving that $\lim_{n\to \infty}P(A_n)$ exists and $\lim_{n\to \infty}P(A_n) =P(\lim\sup A_n)$
Again only for those interested in going to the heart of the matter, I mention that the faulty step in your solution is to believe that $$\limsup x_n+\limsup y_n\leqslant\limsup x_n+y_n,$$ and to use this (wrong) inequality for $x_n=P(A_n)$ and $y_n=-P(A_{n+1})$.