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May
17
revised Show that $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x>-1$ is pointwise convergent.
added 4 characters in body
May
17
comment Series expansion for $x_n=x_{n-1}+\log \left(x_{n-1}\right)$
I understand that you are asking for coefficients $c^n_k$ such that, for every $n\geqslant0$, if $x_0=x$ is close enough to $1$ and if $x_i=x_{i-1}+\ln x_{î-1}$ for every $1\leqslant i\leqslant n$, then $x_n=\sum\limits_kc^n_k\cdot(x-1)^k$. Right?
May
17
comment Series expansion for $x_n=x_{n-1}+\log \left(x_{n-1}\right)$
Not really, but explaining what it is you are asking, definitely would.
May
17
comment Finding a distribution given a moment generating function
$$\frac{z}{1-z}=\sum_{n\geqslant1}z^n\implies\frac{pe^t}{1-(1-p)e^t}=\sum_{n\geq‌​slant1}p(1-p)^{n-1}e^{nt}$$
May
17
comment Series expansion for $x_n=x_{n-1}+\log \left(x_{n-1}\right)$
Series in powers of $x_0-1$, for each fixed $n$?
May
17
comment Show that $\sum_{n=1}^{\infty} \dfrac{1}{n} \ln \left( 1+\dfrac{x}{n} \right)$, with $x>-1$ is pointwise convergent.
This proves nothing when $x<0$.
May
17
comment Why does normal distribution have the same shape regardless of its parameters?
Because each $X_{\mu,\sigma^2}$ can be realized as $X_{\mu,\sigma^2}=\mu+\sigma X_{0,1}$.
May
16
revised Problem on exchangeable events
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May
16
comment sum of the series and infinity
Using the identity $\sum\limits_{i=1}^ki^5=\frac16k^6+$some terms in $k^5$ for $k=10^n$ shows that the result is $\frac1610^{6n}+$some rest of order $10^{5n}$, thus the leading part $166666\cdots$ of the base-$10$ expansion with $n$ or $n-1$ successive $6$s. Likewise with the exponent $59$ since $\sum\limits_{i=1}^ki^{59}=\frac1{60}k^{60}+$some terms in $k^{59}$.
May
16
comment Radius of convergence of $\sum_{n\geq 0}a_{n}x^{n}$.
Hint: If there exists some finite and positive $c$ and $C$ such that $c\le|a_n|\le C$ for every $n$ then the radius of convergence of $\sum\limits_na_nz^n$ is $1$.
May
16
revised How to prove $f(x_1,\ldots,x_n) = \sum x_i\ln x_i - (\sum x_i )\ln(\sum x_i)$ is convex on R++
edited title
May
16
comment Find the general solution of the differential equation $\left(3y^2+x^2+x+2y+1\right)\cdot y'+2xy+y=0$
Do you think you can carry on with this no-personal-input approach forever?
May
16
comment Brownian motion at infinity
Otherwise, by continuity of the paths of $B$, the maximum $M_\infty$ is finite with positive probability. But $M_\infty\geqslant M_t$ almost surely and, by scaling, $M_t=\sqrt{t}M_1$ in distribution hence $P(M_\infty<\infty)\leqslant P(M_1=0)$. Finally, $M_1=|B_1|$ in distribution hence $P(M_1=0)=0$, QED. (Note that several steps in this proof can be solved differently.)
May
16
comment Conditional expectation, quadratic function, absolute value
@Hagrid The canonical approach is to find some measurable function $h$ such that, for every test function $g$, $$E(Xg(Y))=E(h(Y)g(Y)),$$ then one knows that $E(X\mid Y)=h(Y)$ almost surely. In the present case, $X(\omega)=0$ when $\omega<1/2$ hence one asks that $$\int_{1/2}^12(2u-1)g(1-|2u^2-1|)du=\int_0^1h(1-|2u^2-1|)g(1-|2u^2-1|)du,$$ and the rest is a matter of careful changes of variable in integrals.
May
16
comment find the power of a random process?
Obviously, every $$\lim_{T\to\infty}\frac1{2T}\int_{-\infty}^{\infty}\ldots$$ here and in the question should read $$\lim_{T\to\infty}\frac1{2T}\int_{-T}^T\ldots$$
May
16
comment About a probability space
"May you help me, please?" Actually I would not know how. Any suggestion?
May
16
comment Ehrenfest chain
"I tried to find the solution, but I don't understand how to get it, because the book (Introduction to stochastic processes) says that the answer is P(X1=x)=P(X0=x), but it doesn't make sense to me" Forget for one moment that the book gives you the answer, now everything makes sense hence you can solve the question and find the answer, finally compare it to the book's answer.
May
16
comment Variance of [-X]
"if you write the integral, you get something like -integral of x^2 fx dx = -E[X^2]" No you do not.
May
16
comment Alternating Markov process
Let $a$ denote the probability that the light is on when Alice enters the room and $b$ the probability that the light is on when Bob enters the room, then, conditioning on the state of the room one step of time earlier, one sees that $$b=a\cdot\tfrac12+(1-a)\cdot\tfrac14\qquad a=b\cdot1+(1-b)\cdot\tfrac12$$ hence $$a=\tfrac57\qquad b=\tfrac37.$$
May
16
comment Global maximum and minimum of $f(x,y,z)=xyz$ with the constraint $x^2+2y^2+3z^2=6$ with Lagrange multipliers?
"What for?" Mathematical understanding.