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0

I have $c=\frac{2}{7}$ as well. The area is $0.5c+2c+c=1$ $3.5c=1 \Rightarrow c=\frac{2}{7}$ The pdf is $f(x)=\begin{cases} c\cdot (x-1), \ 1\leq x <2 \\ c , \quad \ 2\leq x <4 \\ c-1/2\cdot c\cdot (x-4), \ 2\leq x <4 \end{cases}$ The slope of the third interval is $\frac{f(x_1)-f(x_2)}{x_1-x_2}=\frac{c-0}{4-6}$ We can use the integral ...

0

Your density can be split in three parts $f_1$, $f_2$ and $f_3$: $f_1(x) \cdot x = -2/7 \cdot x + 2/7 \cdot x^2$ $f_2(x) \cdot x = 2/7 \cdot x$ $f_3(x) \cdot x = 6/7\cdot x -1/7 \cdot x^2$ Then then antiderivatives are: $I_1(x) = -2/14\cdot x^2 + 2/21 \cdot x^3$ $I_2(x) = 2/14 \cdot x^2$ $I_3(x) = 6/14 \cdot x^2 -1/21 \cdot x^3$ And finally the sum ...

0

By independence of $X$ and $Y$ we have $P(X>2) * P(Y>2) = P(X > 2 \wedge Y>2)$. Now let $A$ be the event $X>2 \wedge Y>2$ and $B$ the event $X+Y>4$. It is clear that if we have $A$ then we have $B$. The converse however is not true : $$A \Rightarrow B \\ B \nRightarrow A$$ Therefore $P(B) \geq P(A) \Longleftrightarrow P(X+Y>4) \geq ... 0 Comment. As an example, suppose$X, Y \sim Exp(rate=1)$independently. Computation and plot in R for 30,000 simulated realizations. Based on hint by @NicholasStull. (Also, illustrates subsequent answer by @Zubzub.) x = rexp(30000); y = rexp(30000) mean(x > 2)*mean(y > 2) ## 0.01812118 # Aprx P(X > 2)*P(Y > 2) mean(x + y > 4) ## 0.... 2 Yes. Since independence states that the joint probabilty of any sets$\{X \in A\}\;$and$\{Y \in B\}\;$equals the product of both probabilities. 0 Well, the answer $$\Bbb{E}[\max(x_1+x_3+\ldots x_99, x_2+x_4+\ldots x_{100})]$$ is not the correct answer for the expectation value, it is a bit low. To see this, consider the case of only$4$uniform randoms instead of$100$. Here, the expectation in the proposed answer is$ \Bbb{E}[ \max(x_1+x_3, x_2+x_4)] $and you can work out in your head that this ... 0 I assume it should has zero mean and normal distribution. But the acceleration cannot change rapidly. In signal processing, that amounts to generate a stochastic process with low frequency. The simplest way is a AR(1) (autoregresive process) of the form $$X[n]= \alpha X[n-1] + \beta e[n]$$ where$e[n]$is zero mean white gaussian noise, (typically ... 0 $$a_{n + 1} = a_{n} + 10^{-6}\sigma$$ where$\displaystyle{\sigma}$take values$\displaystyle{\pm 1}$with probability$\displaystyle{1 \over 2}$each. I imagine you are thinking in a computer simulation so you can play by changing the$\displaystyle{\sigma}$prefactor which is actualy$\displaystyle{10^{-6}}$. 0 So basically you are looking for a distribution that changes smoothly. There are many, take the Cauchy distribution with mean$0$and the scale parameter near$2$to$3$1 ... if$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}(... 2 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}]$By linearity of expectation we have expectation of sum = ... 2 The marginal distributions of$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\...

5

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 ...

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 ...

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 ...

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 ...

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) \...

3

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

0

The order of the inner summation is reversed. In other words, it is a substitution $$n_1\ \rightarrow\ n_2-n_1.$$

0

Using the hint, $- \frac{1}{n} \sum_{k=1}^{x\sqrt(n)} k - \frac{1}{n^2}\sum_{k=1}^{x\sqrt(n)-1} k^2 < \ln(\mathbb{P}(X^{(n)} < x\sqrt{n})) < - \frac{1}{n} \sum_{k=1}^{x\sqrt(n)} k$ Evaluating the sum and taking $n \to \infty$. We get, $-x^2/2 - 0 < \ln(\mathbb{P}(X^{(n)} < x\sqrt{n})) < -x^2/2$ Taking the exponential we get, $\... 0 How about this: if$Y=X$(that is, they represent the same random variable). Then$X/Y=1$and$Y$and$1$are uncorrelated. 0 Let$X$be a discrete random variable with probability mass function$p_X : \mathcal{X} \to [0,1]$, where$\mathcal{X}$is a discrete set (possibly countably infinite). Random variable$X$can be thought of as a continuous random variable with the following probability density function $$f_X (x) = \sum_{x_k \in \mathcal{X}} p_X (x_k) \, \delta (x - x_k)$$ ... 1 A. Give an example of two discrete random variables$X$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 ... 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. 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$E(X)=cE(Y)=E(X/Y)E(Y)$. 0 No, that is not true. If a random variable can take any value on the real line, but it is exceedingly likely that said variable will be between$0$and$0.1$, then the pdf will have values that are, if not over, then at least close to$10$in that small interval. As another example, if$X\sim\mathcal{N}(0,\sigma^2)$with$\sigma$very small, then$f_X(0)$... 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$\mathcal{F}_2$... 1 Certainly not. Unless the$X_i$are constant, the probability that$\frac{1}{n} \sum \limits_{i = 1}^n 1_A(X_i)$equals$\Lambda(A)$for all$A \in \mathcal{B}(R)$is$0$for any$n$. You need to rethink what you want to show. 1 Let$R$be any one of the red balls, and imagine that we temporarily paint$R$black, so that there are$n-1$red balls and$m+1$black balls in the urn. What is the probability that$R$is the first black ball picked? There are$(m+1)!$equally likely different orders in which the$m+1$black balls can be picked, and$R$is the first ball in$m!$of those ... 0 For two singular matrices$A=\left[\begin{array}{ll}1&0\\0&0 \end{array}\right]$and$B=\left[\begin{array}{ll}0&0\\0&1 \end{array}\right]$, the sum of them is not singular. 1 $${P\{X_1=k\ \cap X_2=m-k\} \over P\{X_1+X_2=m\}}\ne{P\{X_1=k\ \cap X_2=m-k\} \over P\{X_2=m-k\}}$$ In the denominator who said$X_1=k$? Actually it will be $${P\{X_1=k\ \cap X_2=m-k\} \over P\{X_1+X_2=m\}}={P\{X_1=k\ \cap X_2=m-k\} \over \sum\limits_{i=1}^m P\{X_1=i\cap X_2=m-i\}}$$ Now sure you can do the rest... 1 We will need the conditional density of$Y$given$X$. Notice that $$f_X(x) = \int_0^1 f_{X,Y}(x,y)\,dy = x+\frac{1}{2}.$$ Thus, we have that $$f_{Y|X}(y|x) = \frac{f_{X,Y}(x,y)}{f_X(x)} = \frac{x-y+1}{x+1/2}.$$ Finally, we have $$P(Y\geq 1/4|X = 1/4) = \int_{1/4}^1 f_{Y|X}(y|1/4)\,dy = \int_{1/4}^1\frac{1/4-y+1}{1/4+1/2}\,dy = \frac{5}{8}$$ 3 Let's use definitions. By definition, $$\mathbb{P}\left.\left(Y\geq\frac{1}{4}\right|X=\frac{1}{4}\right)=\int_{1/4}^1 f_{Y|X}(y|1/4)dy,$$ where$f_{Y|X}$is the conditional PDF, which is (by definition) $$f_{Y|X}(y|x)=\frac{f_{X,Y}(x,y)}{f_{X}(x)},$$ where$f_Y$is the marginal PDF of$Y, which is (by definition) $$f_X(x)=\int_0^1 f_{X,Y}(x,y)dy.$$ ... 1 Picking up where you left off: \begin{align} \mathbb{E}[e^{itY_n}]=\prod_{k=1}^n \mathbb{E} \left[e^{it\frac{X_k}{2^k}} \right] &= \prod_{k=1}^n \frac{1+e^{it/2^k}}{2} \\ &= \frac{1}{2^n} \prod_{k=1}^n (e^{it/2^k}+1) \\ &= \frac{1}{2^n} (e^{it/2^{n}}+1) \prod_{k=1}^{n-1} (e^{it/2^k}+1) \\ &= \frac{1}{2^n} (e^{it/2^{n}}+1) \frac{(e^{it/2^{n}}-... 2 IfX$and$Y$are iid then$(X,Y)$has the same joint distribution as$(Y,X)$. Consequently$X-Y$has the same distribution as$Y-X$. Since$X-Y$has the same distribution as its opposite, this means that$X-Y$is symmetric. Independence is necessary: For example if the joint distribution of$(X,Y)$is uniform on the three pairs$(0,1)$,$(1,2)$, and$(2,0)...

1

If you want to continue with your approach, just plug in the definitions of $Y_1:=X_1+2X_2$ and $Y_2:=X_1-2X_2$, multiply out, and collect terms. Then use the fact that $E(X^2)=Var(X) + [E(X)]^2$ for any random variable $X$, and that if $X_1$ and $X_2$ are independent, then $E(X_1X_2)=E(X_1)E(X_2)$.

1

let ${{Y}_{1}}=\sum\limits_{i=1}^{n}{{{a}_{i}}}{{X}_{i}}\,$ and $\,{{Y}_{2}}=\sum\limits_{i=1}^{n}{{{b}_{i}}}{{X}_{i}}$. If $X_1,X_2,\cdots,X_n$ are independent then $$Cov({{Y}_{1}},{{Y}_{2}})=\sum\limits_{i=1}^{n}{{{a}_{i}}{{b}_{i}}\operatorname{var}({{X}_{i}})}$$ as a result $$Cov(Y_1,Y_2)=Var(X_1)-4Var(X_2)=-3\sigma^2$$

3

We show that if $X$ and $Y$ are independent Poisson, and $aX+bY$ is Poisson, then $a=1,b=0$ or $a=0,b=1$, or $a=b=1$. We give an elementary argument. It is nicer, but less elementary, to use moment generating functions. Let $W=aX+bY$. With positive probability we have $X=1$ and $Y=0$. So with positive probability we have that $W=a$. It follows that $a$ is ...

1

The calculation would be a special case of the more general discrete convolution formula. Assuming $X$ and $Y$ are as above, we have that the probability mass function of $aX+bY$ is $$\mathbb{P}(aX+bY=z)= \sum_{i=0}^z \mathbb{P}(aX=i)\mathbb{P}(bY=z-i)$$ This becomes complicated in general because (for $Z \sim Poisson(\lambda)$, $m$ a non-...

1

The ages of $A,B,$ and $C$ may be independent†, but the events of pairwise orders are not.   If you are told that $A$ is one of the two oldest children (because $A$ is older than $C$) it should raise your anticipation that $A$ is also older than $B$. († though, actually, they are not independent if they have the same mother; but that is ...

1

Let the children $A, B, C$ have ages $X, Y, Z$ respectively. For each child pair, $\{A,B\}, \{A,C\}, and \{B,C\}$ there are three age possibilities. For the pair $\{A,B\}$, the possibilities are as follows:$$1\ \ \ \ \ X \gt Y$$ $$2\ \ \ \ \ X = Y$$ $$3\ \ \ \ \ X<Y$$ Note that the distribution holds for the other children pairs. Now to answer the ...

2

Let $Y_n=\sum_{j=1}^n 2^{-j}X_j$, then $Y_n$ is uniformly distributed over the set $$\left\{\sum_{j\in S} 2^{-j} : S\subset\{1,2,\ldots,n\} \right\}.$$ This set has the same cardinality of the power set of $\{1,2,\ldots,n\}$, $2^n$, and so $\mathbb P(Y=y)=2^{-(n+1)}$ for each $y\in E_n$. This implies that $\mathbb P(Y_n=y)\stackrel{n\to\infty}\... 1 The problem is that$P((X\gt Y)\cap(X\gt Z))=P(X\gt Y)P(X\gt Z)=\frac14$isn't right. The probability that$X$is greater than both$Y$and$Z$is just$\frac13$, the probability of one of three equivalent children to be the eldest. Thus the events aren't independent. 2 As has been pointed out in the comments, linearity of expectation does not require independence. Thus, the expected number of duplicates, which is$n-m$plus the expected number of empty bins, is$n-m$plus$m$times the probability for a given bin to be empty:$$E[D]=n-m+m\left(1-\frac1m\right)^n=n+m\left(\mathrm e^{-n/m}-1\right)+O\left(\frac nm\right)=\... 2 Note that while it's certainly much easier to go via the cumulative distribution, your approach works, and uses a fairly typical technique for trigonometric products that you might like knowing. At least I always found it kinda cute... First note that$1 + \exp (ix) = 1 + \cos x + i \sin x = 2 \cos x/2 \exp (ix/2)$by using the half angle formulae. Thus, ... 10 The thing you are misunderstanding is that$Var(X-Y)\not=Var(X)-Var(Y)$The accurate formula is$Var(W)=Var(X-Y)=Var(X)+Var(Y)\$

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