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3

As you have guessed, the answer is NO in general. And a single counter example is enough: take $\{X_n\}$ i.i.d. uniform $(0,1)$ then $|X_n|< n$ for all $n\geq 1$ so that $$ Y_n=X_n\implies\text{Var}{Y_n}=\text{Var}{X_n}=\frac{1}{12}\implies\sum_{n=1}^\infty\frac{\text{Var}{Y_n}}{n}=\frac{1}{12}\sum_{n=1}^\infty\frac{1}{n}=\infty. $$


2

The expected value of the number of purchases until you have obtained the complete collection is $E_n=\sum\limits_{i=1}^{n}\frac{n}{n-i+1}=n\sum\limits_{k=1}^{n}\frac1k$. For reasons of symmetry $E(X_i)=\frac{E_n}{n}=\sum\limits_{k=1}^{n}\frac1k$ for all $i$.


1

Since the function $m'_X$ is continuous, we can find $\delta\lt h$ such that if $|s|\lt 2\delta$, then $m'_X(s)-m'_X(0)\leqslant -\mathbb E(X)/2$ (definition of continuity with $\varepsilon:=-\mathbb E(X)/2\gt 0$). We thus have $$|s|\lt 2\delta\Rightarrow m'_X(s)\leqslant \frac{ \mathbb E(X)}2,$$ hence for $0\lt t\lt\delta$, we have $$\tag{*} ...


1

Model: Set $A:=0$, $B:=50$. Let $\xi_j \sim \frac{1}{6} \delta_{-1} + \frac{1}{3} \delta_0 + \frac{1}{2} \delta_1$, $j \geq 1$, be independent identically distributed random variables. Then $$X_n := k + \sum_{j=1}^n \xi_j$$ is the position of the person after $n$ steps if the person starts at $X_0=k$. We would like to calculate the probability that the ...


1

You got indeed the idea, and your proof looks correct. Some comments: In the second line of your attempt, the expression of the p.d.f. of a uniform distribution in the unit interval is not correct if $t\notin [0,1]$. How do you define the random variable $f(x)$? Do you mean $f(U_1)$? Actually, you don't need to use the weak law of large numbers, but ...


1

The Fisher information is essentially the negative of the expectation of the Hessian matrix, i.e. the matrix of second derivatives, of the log-likelihood. In particular, you have $$l(\alpha,k)=\log\alpha + \alpha \log k - (\alpha + 1) \log x$$ from which you compute the second-order derivatives to create a $2 \times 2$ matrix, which you take the expectation ...


1

Define $$A_n := \left\{ \left| \frac{X_n}{n} \right| \geq 1 \right\}.$$ The set $$A := \limsup_{n \to \infty} A_n$$ satisfies, by the Borel-Cantelli lemma, $\mathbb{P}(A)>0$ and $\frac{X_n(\omega)}{n}$ does not converge to $0$ for each $\omega \in A$.


1

Hint: Find first $P(X+Y=k|X=x)$, and then use this to find using Bayes's theorem $P(X=x|X+Y=k)$. Also note that you need $X,Y$ to be independent otherwise things won't work. Now, use the definition of conditional expectation to obtain $E(X|X+Y=k)$.



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