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Let $a = \mathbb E[X]$ and $b = \mathbb E[Y]$; you want to prove that $|a^2 - b^2| \leq (a - b)^2$. The expected value formulation is sort of a distraction here. With this formulation, you can fairly easily find a counterexample to show that the statement is false. HINT: WOLOG, suppose $a > b$. Then the left side is $(a + b)(a-b)$ and the right side is $(... 2 Looking at RHS of equality: $$\mathbb EX=\sum_ix_iP(x_i)$$ we observe that a term$x_iP(x_i)$is only relevant if: $$x_iP(x_i)\neq0$$ That will be the case here if and only if$x_i=1$and in that case: $$P(x_i)=P(1)=\Pr(X=1)=\Pr(N\leq5)=\frac5{10}=\frac12$$ So we find:$$\mathbb EX=1\cdot\frac12=\frac12$$You can find$\mathbb EYon a similar way. 2 \begin{align} P(Z=k) &= e^{-k}-e^{-k-1}\\ &= e^{-k}\left(1-e^{-1}\right) \end{align} It follows Geometric distribution which start from0$. 2 The brownian motion$B_t$has a symmetric distribution arround 0 (more precisely, a centered Gaussian). Since$sin$is an odd function, then$\mathbb{E}[\sin(B_t)] = 0$for all$t$. To see this, since$-B_t$has the same distribution as$B_t$, we have that $$\mathbb{E}[\sin(B_t)] = \mathbb{E}[\sin(-B_t)] = -\mathbb{E}[\sin(B_t)]$$ which gives$\mathbb{E}[\...

2

Use the Stolz–Cesàro theorem (an analogy of the L'hospital theorem for sequences): $$\lim_{N\to\infty}\frac{\sum_{n=1}^Na_n}{N}=\lim_{N\to\infty}\frac{\sum_{n=1}^{N+1}a_n-\sum_{n=1}^Na_n}{(N+1)-N}=\lim_{N\to\infty}a_{N+1}=0$$

2

Please see if the below helps: While you cannot do is to escape CLT, what you can do is to reduce the variance of the individual random variables you add. Translating your requirement, you want a 36% CI that the sum of 100 RV's are between 89 and 91. This translates to a z value between $-0.47$ and $0.47$. Therefore the standard deviation of $100$ variables ...

1

The two rv's are both bernulli with parameter 0.5 thus their expectation is $0\times 0.5 + 1\times 0.5 =0.5$ This because the rv that describes the event $N\leq 5$ can take only the values 0 and 1 (false or true) with probability 0.5 as you have 5 favourable events among 10 equiprobable ones Similar reasoning for the other rv

1

Let $M_X(t)$ denote $X$'s moment-generating function, $E[e^{tX}]$. You want$$E[X^2e^{2X}]-(E[Xe^X])^2=M_X^{\prime\prime}(2)-(M_X^\prime(1))^2.$$Famously, $M_X(t)=(q+pe^t)^n$ with $q:=1-p$ (proof is an exercise). The rest is a calculus exercise.

1

It depends on the variance of the population. If it turns out that conversions from A have a higher variance than conversions from B, then to achieve equal bounds on each estimate, you'll need more tests of the former than the latter. It seems that in proportion testing people often assume the maximum variance for establishing the number of tests to perform (...

1

Let's say we had some random variables $X$ and $Y$ such that $P(X > a \cap Y > b) = \frac{a}{b}P(X > a)P(Y > b)$ for $b > a > 0$, else $P(X > a)P(Y > b).$ Clearly, letting $a = b$ would satisfy the first criterion: $$P(X > a | Y > a) = \frac{P(X > a \cap Y > a)}{P(Y > a)} = \frac{\frac{a}{a}P(X>a)P(Y>a)}{P(Y>a)}... 1 As the sample sizes increases, density estimators do tend to come ever closer to the distribution of the population from which the sample was randomly samples. (However, depending on the kernel and bandwidth used, ECDFs can be computationally intricate.) Here is an illustration using samples of size 50, 500, 5000 from a gamma distribution with shape 5 and ... 1 To Prove: \mathbb{P}\{T_A>kN\}\leq \left(1-\frac{1}{2d}\right)^k. First we do this for 1 dimensional case and then extend Assumption: Let the points in A be contiguous. Further, let us start from the center of this interval. Any other case will have a higher probability that we hit the edges of the wall before N steps, so this is a conservative ... 1 This is one of my favorite proofs about Cesaro means. There is a quicker proof here but for some reason I prefer the proof I give below, and I thought I'd share it just in case. Proposition Let (a_n)_{n\in\mathbb{N}} be a sequence of real numbers converging to a\in\mathbb{R}. Define$$C_n=\frac{1}{n}\sum_{k=1}^{n}a_k$$Then$$\lim_{n\to\infty}C_n=a$$... 1 You can also apply the CLT. The sum of a large number of random variables tends towards a normal distribution. Here you are taking the product of a large number of random variables. Take log on both sides and apply the central limit theorem. Then Y\sim log(X) has a normal distribution, and therefore X\sim \text{lognormal} PS: Your working looks correct ... 1 Suppose the patients are given the pills in the order H_1H_2H_3H_4D_1D_2D_3A_1A_2O_1 where H_i represents the person who has to take the pill for hypertension, D_i for diabetes, A_i for arrhythmia and O_1 for hypotension. Now, probability that H_1 gets the right pill is 4/10, then H_2 gets the right pill is 3/9 so in this way, the ... 1 For (1), choose a random integer between 1 and 100. Call the result A, B or C using the intervals [1,90], [91,98], [99,100]. I know that's not the algorithm you ask for. Intuition suggests that you cannot guarantee (2) with any algorithm. What coin tossing algorithm would you use to assure between 49 and 51 heads with predetermined ... 1 It is not quite correct. The total number of equally probable ways of choosing ten from 100 is 100C10 or {100 \choose 10} Of these, the number of ways of choosing A and B and eight others ten from 100 is 2C2 \times 98C8 or {2 \choose 2}{98 \choose 8}. We can ignore the {2 \choose 2}=1 term and so the probability is$$\frac{98 \choose 8}{...

1

No, your thoughts aren't correct. For the first part, $\dfrac{\binom{98}8\binom2 2}{\binom{100}{10}}$ For the second part, which you should now do by yourself. there are only $98$ people to select from for the numerator.

1

You are computing the probability that there are no $6$s in the first $x-1$ rolls, but they want to compute the probability that faces $1,2,3,4,5$ each appear at least once in the first $x-1$ rolls.

1

One good way is to try to convert this to a more standard linear recursion. Letting $$a_n=6(a_{n-1}+1)=6a_{n-1}+6$$ We rewrite this as $$6=a_n-6a_{n-1}=a_{n-1}-6a_{n-2}$$ Whence $$a_n=6a_{n-1}+a_{n-1}-6a_{n-2}=7a_{n-1}-6a_{n-2}$$ and this can now be solved by standard means, yielding $$\boxed{a_n=\frac 65\times \left(6^n-1\right)}$$

1

$$E[Y] = E[\pi R^2] = \pi E[R^2] = \pi \cdot \int_{0}^{c}t^2f_R{t} = \pi \cdot \int_{0}^{c}t^2\cdot \frac{1}{c} = \pi \cdot \int_{0}^{c}t^2\cdot \frac{1}{c} = \pi \frac{c^2}{3}$$ Used: Linearity of expected value, Linearity of integrals,Smooth transformation of continous Random variable

1

CDF when $x$ and $y$ are uniform $(0,1).$ $$P(Z\leq z)= \begin{cases} \frac{z^2}{2}& 0< z \leq 1\\ 1-\frac{(2-z)^2}{2}& 1<z<2. \end{cases}$$ Taking the derivative, the PDF is $$f(z)= \begin{cases} z & 0<z\leq1\\ 2-z & 1<z<2. \end{cases}$$ If $X$ and $Y$ are uniform $(\theta,\theta+1),$ then $X-\theta$ and $Y-\theta$ are ...

1

Here's a hint: Let $A_i$ denote the event that we don't draw $F_i$ in any of our six draws, and then use inclusion-exclusion to compute $P(A_1 \cup A_2 \cup A_3)$. Then consider how this relates to the probability you want to find.

1

You cannot use the same $Z$ for $(B_t)$ and $(W_t)$ because they are independent. So you have to evaluate $P(\sqrt {\frac {e^{2}-1} 2} Z >\sqrt {\frac 1 {24}} W)$ where $Z,W$ are i.i.d. $N(0,1)$. Can you finish?

1

Your approach is ok, though the sum should begin in $k=1$ (or $k=2$). I don't think the expression has a nice closed form. Noticing that the mean of the variable is $\lambda = 10$, and that the approximation $(1-1/n)^n \approx e^{-1}$ should work reasonably around $n\approx 10$, then the function $g(n)= n (1-1/n)^n$ is approximately linear around the mean. ...

1

The correct answer is $0.02$, and your second approach is correct. Here is how you can make your first approach work: You should multiply 2/99 (probability of ending at the second round given you did not end in the first round -- see below) by 99/100 (probability of not ending in the first round) and you will get the same answer. $$P(B|A) = \frac{49}{99} \... 1 The second calculation is correct. The mistake in the first calculation is that \Pr(B|A)is not \frac2{100} but \frac2{99} Once we know that 50 is not the number, only 99 possibilities remain. If you want to do it by formula,$$\Pr(B|A)=\frac{\Pr(B\cap A)}{\Pr(A)}=\frac{\Pr(B)}{\Pr(A)}=\frac{2/100}{99/100}=\frac2{99}

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