# Tag Info

6

One simple case where your identity is true: Let $Y$ be some nonconstant and positive RV, and let $X:=cY$ for some nonzero $c$. Then $E(X)=cE(Y)=E(X/Y)E(Y)$.

4

Using Stirling numbers we get $$\frac{1}{25^{10}} \sum_{q=1}^{10} q\times {25\choose q} \times {10\brace q} \times q!$$ which gives the expectation $${\frac {31964050675249}{3814697265625}} \approx 8.379184100.$$ What we have done here is choose the $q$ different values from the $25$ different ones where we can obviously represent at most ten ...

3

Try sample space $\{1,\ldots,10\}$ with equal probabilities, $X(i) = i$ and $Y=X+1$ when $X<10$, $1$ when $X=10$. If $X$ and $Y$ are independent, $X=Y=i$ has positive probability whenever $X=i$ and $Y=i$ do.

3

See https://en.wikipedia.org/wiki/Renewal_theory#The_elementary_renewal_theorem The elementary renewal theorem

2

Let $\Omega=\{1,2\}^3$ with the uniform probability law, and let $X_i$ be the projection onto the $i$th coordinate. Then $\mathcal{F}_1=\sigma(X_1,X_2)$ contains the event $$\{X_1=1\}=\{(1,1,1),(1,1,2),(1,2,1),(1,2,2)\}$$ Suppose for contradiction that $\mathcal{F}_2=\sigma(\frac{X_1}{S},\frac{X_2}{S})$ also contains this event. Since $\mathcal{F}_2$ ...

2

We get the same answer for any continuously distributed random variable. For let $M$ be a median (medians need not be unique). The probability that $X_i$ is $\gt M$ is $\frac{1}{2}$. The probability that the minimum is $\gt M$ is therefore $\left(\frac{1}{2}\right)^3$. Remark: Things can break down if the distribution is not continuous. For example, let us ...

2

To make things slightly simpler, say there are $n$ balls in the box. Tge expected value of the first ball is (n + 1)/2. Suppose we've drawn k balls already and the maximum so far is M. The expected value of the next maximum is $$E(M_{next}) = M^2/n + (M+1)/n + (M+2)/n + \ldots + n/n$$ This gives you a recurrence relation for $E(M_k)$ that you can ...

2

We give a very weak result: sometimes the only functions $g$ are the trivial ones. Let $X$ take on values $0$ and $1$, and let $\Pr(X=1)=p$, where $p$ is transcendental. Let $g(0,0)=a$, $g(0,1)=b$, $g(1,0)=c$, $g(1,1)=d$. Here $a,b,c,d$ only take on values $0$ and $1$. The corresponding probabilities are $(1-p)^2, p(1-p), p(1-p),p^2$. There are two ways the ...

2

The marginal distributions of $X_1$ and $X_2$ can be obtained by "integrating out" the other variable. For instance, $$f_{X_1}(x_1)=\int f_X(x_1,x_2)\;dx_2=\int_0^1\Big(\frac{2}{3}x_1+\frac{4}{3}x_1x_2+\frac{2}{3}x_2\Big)\;dx_2=\frac{4}{3}x_1+\frac{1}{3}$$ if $0\leq x_1\leq 1$. The same argument shows that $f_{X_2}(x_2)=\frac{4}{3}x_2+\frac{1}{3}$ for $0\... 2 Yes. Since independence states that the joint probabilty of any sets$\{X \in A\}\;$and$\{Y \in B\}\;$equals the product of both probabilities. 1 ... if$R_1$and$R_2$both denote a matrix we get$R_1$and$R_2$should be vectors. Then$Var(R_1+R_2)=\Sigma_1+\Sigma_2+2\rho_{12}\sqrt{\Sigma_1} \sqrt{ \Sigma_2}$where$\Sigma_i$denotes the variance matrix of$R_1$and$R_2$respectively. 1 When working with multivariate variances and covariances, it's good to keep this notational advice in mind. I'll stick with your notation and use$\operatorname{Var}(R)$to denote the (co)variance matrix of the random vector$R$, i.e.$\operatorname{Var}(R)=\operatorname{cov}(R,R). Then \begin{align} \operatorname{Var}(R_1+R_2)&=\operatorname{cov}(... 1 LetX_i$be an indicator random variable$=1$if the$i^{th}$coupon is present, and$=0$otherwise. Then$P[i^{th}$coupon is present$] = [1 - (\frac{24}{25})^{10}]$Now the expectation of an indicator r.v. is just the probability of the event it indicates, so$E[X_i] = [1 - (\frac{24}{25})^{10}]$By linearity of expectation we have expectation of sum = ... 1 Given that all$X_i$are at most$s$, all values of$X_1$from$1$to$s$are equiprobable, so the expected value of$X_1$is$\frac{s+1}2$, and the expected value of$T_1$is$2\cdot\frac{s+1}2-1=s$. The event that the maximum is$t$is the event that all$X_i$are at most$t$minus the event that all$X_i$are at most$t-1, and the probabilities for those ... 1 Yes, in a finite case like this the expected value and variance exist; you don't need to prove that in detail. You can calculate the expected value from the expression you wrote, but quite often, including in this case, it's easier to calculate it like this: \begin{align} E(M)&=\sum_{m=0}^{48}P(M\gt m) \\ &= \sum_{m=0}^{48}\left(1-P(M\le m)\right) \... 1 A. Give an example of two discrete random variablesX$and$Y$on the same sample space such that$X$and$Y$have the same distribution, with support {1, 2, . . . , 10}, but the event$X = Y$never occurs. An example similar to Robert Israel’s: Choose any symmetric probability distribution for$X$; i.e.,$P(X=1)=P(X=10)$,$P(X=2)=P(X=9)\$, etc.  A ...

Only top voted, non community-wiki answers of a minimum length are eligible