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1d
comment Using characteristic functions to establish convergence
Would you mind elaborating on the parts of my comment which elude you please?
1d
comment Expectation problems in probability.
Did you get the hint?
1d
comment Using characteristic functions to establish convergence
$$e^z-1-z\sim\tfrac12z^2\implies\lambda\cdot(e^{itb(\lambda)}-1)-it \lambda b(\lambda)=\lambda\cdot(e^{itb(\lambda)}-1-it b(\lambda))\sim\lambda\cdot\left(-\tfrac12t^2b^2(\lambda)\right)$$
1d
comment Expectation problems in probability.
Please read: Indeed it does. (You know what, some of us can read the statement of an exercise...)
1d
comment Interpretation of composite of random variable
Thus, no ambiguity left.
1d
comment solve the inhomogeneous system
-1. Please read: How do I find the particular solution? Keyword: How.
1d
comment Expectation problems in probability.
Indeed it does. Hint: reread the definition of $H_i^k$ (no idea what your H's means).
1d
comment How to prove that the sum of a convergent geometric series of the form $1 + r + r^2 … + r^n > 1/2$?
But if $r=-\frac34$, this is false for $n=1$ and $n=3$.
1d
comment Interpretation of composite of random variable
Assuming that $f\leqslant c$, obviously not. Assuming that $|f|\leqslant c$, obviously yes.
1d
comment Find the value of $\sum^{n-1}_{m=1}\left(\frac{1}{n-m}+\frac{1}{n+m}\right)$.
The $\psi^{(0)}$ formula is a mere rewriting, mainly useless. You might want to use $$\sum^{n-1}_{m=1}\left(\frac{1}{n-m}+\frac{1}{n+m}\right)=\sum^{n-1}_{k=1}\frac‌​{1}{k}+\sum^{2n-1}_{i=n+1}\frac{1}{i}=\sum^{2n-1}_{m=1}\frac{1}{m}-\frac{1}{n}-\f‌​rac{1}{n+1}.$$
1d
comment Expectation problems in probability.
Then forget the events $T_i^k$ (and replace $3^{-k}$ in your answer by $(2/3)^k$).
1d
comment Any idea on this problem $\lim \limits_{x\to\infty}f(x)=0$
This might be one of the questions in analysis which is most often asked on the site.
1d
comment Interpretation of composite of random variable
Yes. $ $ $ $ $ $
1d
comment Expectation problems in probability.
Hint: Let $H_i^k=\{X_{i-k}=\cdots=X_{i-1}=X_i=H\}$ and $T_i^k=\{X_{i-k}=\cdots=X_{i-1}=X_i=H\}$ then $$N_k=\sum_{i=k+1}^n\mathbf 1_{H_i^k}+\mathbf 1_{T_i^k}.$$ Since $P(H_i^k)=(1-p)^k$ and $P(T_i^k)=p^k$ with $p=\frac13$, this almost sure identity yields $$E(N_k)=(n-k+1)(p^k+(1-p)^k).$$
1d
comment Prove that if $A \mathbin{\triangle} C = B \mathbin{\triangle} C$, then $A = B$
Full proof: $$(A\triangle C)\triangle C=A$$
2d
comment What is the decomposition of $x^4+x^3+x^2+x+1$.
What @Jyrki said.
2d
comment What is the decomposition of $x^4+x^3+x^2+x+1$.
Hint: Assuming you ask for the decomposition over the reals, note that this is $x^2(u^2+u-1)$ where $u=x+1/x$. Now, factor $u^2+u-1$.
2d
revised If $X_n$ is Geo$(\lambda/n)$, does $X_n$ converge in distribution?
deleted 22 characters in body; edited title
2d
comment If $X_n$ is Geo$(\lambda/n)$, does $X_n$ converge in distribution?
Indeed $P(X_n\leqslant x)\to0$ for every $x$ hence $(X_n)$ does not converge in distribution. (By the way, one can couple $(X_n)$ in such a way that $X_n\to\infty$ almost surely, which forbids convergence in distribution. Last remark: $X_n/n$ does converge in distribution, to a well-known continuous distribution...)
2d
revised Coupon Collector's Problem — Expected Value of each item
added 66 characters in body