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3

Note that $E[X+Y]=E[X]+E[Y]$ holds in full generality, even if $X$ and $Y$ are not mutually independent. Proofs of linearity of expectation do not assume independence of $X$ and $Y$. Here's one for example. In other words you do not need to impose that restriction.


3

If $M = \min(X_{1}, . . . , X_{n})$, it can be shown that $M$ has an exponential distribution with parameter $n \lambda_0$. So $E(M)=\frac 1 {n \lambda_0}$ Since $T=nM$, it follows that $E(T)=E(nM)=nE(M)=\frac 1 {\lambda_0}$ Details can be found here: Distribution of the minimum of exponential random variables


2

I think the problem is that the number of attempts that can be used in a numerical simulation $n$ is finite. Notice this: if $Y_n=\frac{S_n}{\sqrt{2n\log\log n}}$, by properties of random walk we know $\mathbb{E}[Y_n]=\frac{\mathbb{E}[S_n]}{\sqrt{2n\log\log n}}=0$ and $$ Var[Y_n]=\frac{Var[S_n]}{2n\log\log n}=\frac{n}{2n\log\log n}=\frac{1}{2\log\log n}\to ...


2

I encourage you to work through the details of your own approach. Here is my alternative. After the first drawing (a pair of balls), we have exactly two balls (out of five possible) that have been drawn (and we can never get fewer than this in the end). The second drawing can affect which balls have been seen. There are three possible outcomes as far as ...


2

Think of 75% as the probability that the stock goes up, i.e. $0.75$. Then, the trader gains 100 with probability $0.75$, loses 200 with probability $1-0.75=0.25$. On expectation, what is his gain? More formally: let $X$ be the random variable representing his gain. Then, $\mathbb{P}\{X=100\} = 0.75$, and $\mathbb{P}\{X=-100\} = 0.25$. You are asked to ...


2

There are $32^4$ possible outcomes for $(a,b,c,d)$, since each one has $32$ outcomes. a) If none of them are equal, then outcomes for $(a,b,c,d)$ are $32\times 31\times 30\times 29$. Thus, the probability is $\frac{32\times 31\times 30\times 29}{32^4}=\frac{13485}{16384}=0.823$ b) Select the common number: $32$ outcomes. Select the places: $\binom{4}{2}$. ...


2

By the Borel-Cantelli lemma, if the series $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} $$ converges for each $\varepsilon>0$, $Y_n\to0$ almost surely as $n\to\infty$. Using Chebyshev's inequality, $$ \sum_{n=1}^\infty P\{|Y_n|>\varepsilon\} \le\frac1{\varepsilon^2}\sum_{n=1}^\infty\operatorname E|Y_n|^2 $$ since $\operatorname EY_n=0$. By ...


2

To simplify the subscripts, I'm going to consider just the first few letters the monkey types. Add a variable $k$ to all my subscripts if you want to apply the reasoning at an arbitrary point in the string. It is true that $E(X_1) = 26^{-5}$ and also that $E(X_2) = 26^{-5}$. We just have to cycle through the $26^6$ equally-likely possibilities for the first ...


2

For (1): show you can have, for instance, $\operatorname{var} Y = 0$ and $\operatorname{var} X >0$, or $\operatorname{var} Y >0$ and $\operatorname{var} X = 0$. For (2): Show you can have $\mathbb{E} X = 0$ and $\mathbb{E} Y > 0$, or $\mathbb{E} X = 0$ and $\mathbb{E} Y < 0$. For (3): You should see the relation with (1). For (4): Show you ...


1

Hint: Think about different variations of what X and Y are allowed to be given our assumptions. In particular, compare cases when $X$ is either identically zero or uniformly distributed, and $Y$ is identically $-2$ or $+2$. See if you can generate counter examples by using these.


1

To construct counterexamples: Let $Y\equiv 2$ and $X \sim U(-1,1)$. Then $Var(X)>0=Var(Y)$. Let $Y \equiv -2$ and $X\sim U(-1,1)$ or simply $X\equiv 0$. Then $E[X]=0>-2=E[Y]$. Let $Y\in\{-2,2\}$ with $P(Y=-2)=p=1-P(Y=2)$ and $p>0$ and $X\equiv 0$. Then $Var(Y)>0=Var(X)$. Let $Y\equiv2$ and $X\sim U(-1,1)$. Then median $Y=2$ and median $X=0$. ...


1

Where in this proof have we imposed that restriction? Your observation that $X_i=1 \implies \sum_{n=1}^4X_{i+n}=0$ restricts $X_{i+1}\cdots X_{i+4}$ is certainly valid. There is no need to take this restriction into consideration, because it is self-imposing:$$X_i=1 \implies \sum_{n=1}^4X_{i-n}=0$$ Edit: $E[X_{i+1}]\cdots E[X_{i+4}]$ are not ...


1

This is not an answer to the question, so this may not be strictly speaking quite welcome here, but I think that any putative proof that the question's expectation formula is not valid should account for this data somehow, and so should any alternative expectation formula for this experiment. The formula in the question seems to match the simulation results ...


1

Note that $$P(X^n\in C_n(t))=P(p(X^n)\ge 2^{-nt})=P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le t\right)$$ Now, it follows from WLLN, $$\lim_{n\to \infty}P\left(-\frac{1}{n}\sum_{1}^n \lg p(X_i)\le H(X)+\epsilon\right)=1\ \forall \epsilon>0$$ Thus, to have $\lim_{n\to \infty}P[X^n\in C_n(t)]\to 1$, we must have $t\ge \epsilon+H(X)\ \forall ...


1

You say "In both cases it seems advisable to bet everything". But your intuition is not correct. In fact, if you bet almost everything and lose, the log of your final amount of money approaches negative infinity, but if you win it approaches a finite value. So betting everything actually gives you infinitely negative expected value! To see where the $ ...


1

It is correct, but can be further simplified by considering the order statistics of the sample. That is to say, if $$\boldsymbol x = (x_1, \ldots, x_n)$$ is the sample, then consider $$x_{(1)} = \min_i x_i, \quad x_{(n)} = \max_i x_i,$$ the first and last order statistics which are equal to the smallest and largest observations in the sample, respectively. ...


1

Note that $X_n\gt 0$ surely for each $n$, hence by definition of $g$, we have $g\left(X_n\right)=1$. The random variable $X$ is not specified but we can imagine that this is the limit in probability of the sequence $\left(X_n\right)_{n\geqslant 1}$. Since $X_n\to 0$ in probability and $g(0)=0$, we do not have $g(X_n)\to g(X)$ in probability (the problem ...


1

Hint: use the Borel—Cantelli lemma. In more detail: fix any $\varepsilon > 0$, and let $A_n=A_n(\varepsilon)$ be the event $\{ \lvert Y_n\rvert > \varepsilon \}$. By Borel—Cantelli, to show that $Y_n \xrightarrow[n\to\infty]{\rm a.s.} 0$, it is sufficient to show $$ \sum_{n=1}^\infty \mathbb{P} A_n < \infty. $$ In even more detail: (place your ...


1

If $X=Y$ then 1,2,3,4 would be correct. In particular $X-Y=0$ with probability $1$ If $X$ and $Y$ are independent then 1,2,4 are correct but 3 might not be. For example, suppose $X$ takes the value $1$ with probability $0.4$ and the value $0$ with probability $0.6$ But if $X$ and $Y$ have a more complicated relationship then there is little you can say ...


1

Consider a sequence of four observations, one from each random variable: $$(x_1,x_2,x_3,x_4).$$ There are $32^4 = 1{,}048{,}576$ such sequences, each of which is equally likely. For a), there are 32 choices for $x_1$, leaving 31 choices for $x_2$, leaving 30 choices for $x_3$, leaving 29 choices for $x_4$. By the multiplication principle, there are $32 ...


1

Take your geometric random variable distributed as $p_k=\mathbb P (X=k)=p(1-p)^{k}$ for $k\in\{0,1,2,\dots\}$. Note that your entropy is: $$ H(X)=-\log p-\frac{1-p}{p}\log(1-p). $$ And consider another random variable $Y$ distributed as $q_k=\mathbb P (Y=k)$ such that: $$ \frac{1-p}{p}=\mathbb E(X)=\mathbb E(Y)=\sum_{i=0}^\infty i q_i. $$ We use following ...


1

The following approach is feasible only because the numbers are very small. Without loss of generality we may assume that on the first draw we got two specific balls, say #4 and #5. Then our desired event can happen in two ways: (i) We get two new balls in Draw 2, and the missing one in Draw 3, or (ii) We get one of balls #4 or #5 and a new one in Draw 2, ...


1

You've neglected the possibility of ties and are over counting events where multiple dice equal $y$. You wish to calculate the probability that two dice are $x$ and $y$ and none of the remaining die are higher than $y$. There are two cases to consider. When $x=y$ and when $x>y$ When $x=y$ you want the probability that all dice are at most $x$, ...


1

Mostly correct. In the case of $a=0$ then $F_Y(t) = \begin{cases} 0 &: t<b\\1 & :t\geq b\end{cases}$ because, as you reasoned, $Y$ would be a deterministic random variable (a single massive point at $b$). So the probability of $Y$ being at most $t$ is zero if $t<b$ and one if $t\ge b$. I don't see why if $X$ is a discrete random ...


1

$$ \mathbb{E}[Z^2\mid M] = \mathbb{E}\left[\left(\sum_{k=1}^N Y_k\right)^2\right] = \mathbb{E}\left[\sum_{k=1}^N \sum_{\ell=1}^N Y_k Y_\ell\right] = \sum_{k=1}^N \sum_{\ell=1}^N \mathbb{E}\left[Y_k Y_\ell\right] $$ by linearity of expectation. Now, $Y_k$ and $Y_\ell$ are independent if, and only if, $k\neq \ell$, so $$\begin{align} \mathbb{E}[Z^2\mid M] ...



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