# Tag Info

3

There is a much more general result which is true. Proposition Let $X$ and $Y$ be independent, real-valued, integrable and centered random variables. Let $\varphi$ be a non-negative convex function. Then: $$\mathbb{E} (\varphi (X+Y)) \geq \mathbb{E} (\varphi (X)).$$ Proof The main tool is a conditional version of Jensen's inequality ...

3

Considering $$I=\int e^{-\frac c x} dx$$ make a change of variable $-\frac c x=y$ that is to say $x=-\frac c y$, $dx=\frac c {y^2}dy$ to make $$I=c\int \frac{ e^y}{y^2}dy$$ Now, integration by parts $$u=e^y\quad du=e^y dy\quad dv=\frac{ dy}{y^2}\quad v=-\frac 1 y$$This gives I=c\left(-\frac {e^y} y+\int \frac{ e^y}{y}dy\right)=c ... 3 Let \bar x =(x_1+\cdots+x_n)/n be the sample mean. Then \begin{align} \sum_{i=1}^n (x_i - \mu)^2 & = \sum_{i=1}^n ((x_i-\bar x) + (\bar x - \mu))^2 \\[10pt] & = \sum_{i=1}^n \Big( (x_i-\bar x)^2 + 2(x_i-\bar x)(\bar x-\mu) + (\bar x - \mu)^2) \\[10pt] & = \left( \sum_{i=1}^n (x_i-\bar x)^2 \right) + \left( 2 (\bar x - \mu) ... 3 To find the p.d.f of the ratio \frac{Y}{X+Y}, let us first write its c.d.f. Since X and Y are always positive, their ratio is also positive and, therefore, for 0\leq t\lt1 we can write:  P\left(\frac{Y}{X+Y}\leq t\right)=P\left(Y\leq \frac{t}{1-t}X\right)=\int_{0}^{\infty }\left(\int_{0}^{\frac{t}{1-t}x}f_{X}(x)f_{Y}(y)dy\right)dx  as ... 2 You have essentially rediscovered the theorem that says that any faithful compression algorithm must increase the size of some input files in order to compress others - one hopes, most others. Google lossless compression pigeonhole principle for several links. https://en.wikipedia.org/wiki/Pigeonhole_principle#Uses_and_applications ... 2 Let's assume that "is certain of the correct answer" = "knows the correct answer". Without assuming that, this "certainty" would be equivalent to "guessing randomly". What is the probability that the student correctly answers question #1? Split it into disjoint events, and add up their probabilities:0.6+(1-0.6)\cdot\frac14=0.7$$What is ... 2 Inclusion/Exclusion will work. It is easier to find the probability of the complement, that is, the probability that X and Y are both between 0 and 2. So our required probability is$$1-\int_0^2\int_0^2 \frac{1}{54}(x^2+y^2)\,dy\,dx.$$Remark: Inclusion/Exclusion yields$$\int_{x=0}^3\int_{y=2}^3 k(x^2+y^2)\,dy\,dx+\int_{y=0}^3\int_{x=2}^3 ...

2

$$E[Y_1] = \int_0^1 2y_1^2dy_1 = \frac23$$ $$E[Y_2] = \int_0^1 2y_2^2dy_2 = \frac23$$

2

The problem may be formulated differently, which might hint to a solution: what is the probability that you get no $6$, and at least one $5$? The probability of none of the dice showing a $6$ is $(5/6)^3$. Given that you have gotten no $6$, the probability of getting at least one $5$ is the opposite of the probability of no $5$. The probability of no $5$ ...

2

The only way this makes sense if it is with replacement. That is each contestant can chose any number. Here is why: If it were without replacement it would not make sense to assign a new winning number after each try. This makes sense in a scenario with replacement so that each contestant still has a $1$ in $50$ chance of winning, as earlier attempts do ...

2

Suppose $10,000$ people are tested (to avoid decimals) Of these, $32\%$ or $3200$ have the disease, of which $97\%$ or $3104$ test positive $6800$ don't have the disease, of which $12\%$ or $816$ test positive Thus P(have the disease | test positive) $= \dfrac{3104}{3104+816}$

1

In general what you want to do is calculate the money per hour for taking the toll or in your notation $p/t$. Then you want to compare that value to the value of your time. If the money per hour calculated is less than the value of your time then it is worth taking, otherwise it is not. In your case the money per hour for the toll is $p/t=2/(1/4)=8$. So ...

1

Define $R$ to be the event of a positive test result, $R^c$ to be the event of a negative test result, $D$ to be the event that the person is diseased, $D^c$ to be the event that the person is not diseased. We are given that $P(R|D)=0.97$ and $P(R^c|D^c)=0.88$, i.e. when a person does have the disease, the test result is positive, and when he doesn't, the ...

1

The problem of estimating the true weights $m_i$ is best studied using a matrix formulation. Consider the "weighing matrix" $$A = \begin{pmatrix} 1&1&0 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} \, .$$ Let $m$ be the vector of true weights. You have two observations per combination of objects, namely $b_1 = Am + \xi_1$ and $b_2 = A m ... 1 Where$\mathbf{1}$is the indicator function, $$f(x)=\frac{1}{400} (x-80) \mathbf{1}_{80\leq x\leq 100}+\frac{1}{400} (120-x) \mathbf{1}_{100\leq x\leq 120}$$ Moments: $$E(X^n)=\frac{2^{2 n+1} 5^n \left(2^{2 n+3}+2^{n+1} 3^{n+2}-5^{n+2}\right)}{(n+1) (n+2)}$$ Mean:$n=1$,$E(X)=100$So the variance$V(X) =E(X^2)-E(X)^2=\frac{200}{3}$1 The expectation is linear - that is,$E(aX + bY) = aE(X) + bE(Y)$for all$a,b\in \mathbb{R}$and$X,Y$. In particular $$E(W) = E(5) - \frac{1}{2}E(Y) = 5 - \frac{1}{2}E(Y)$$ However, the variance is not linear (unless the random variables are uncorrelated). Instead, it satisfies $$V(aX + b) = a^2V(X)$$ for all$a,b\in \mathbb{R}$and$X$. Thus, we ... 1 (1.) Let$s$be defined as$S_n = x_1+\dotsb+x_n$. Then the distribution of$s = S_n$is $$P(S_n = k) = \Sigma^{(k)}\prod_{i=1}^np_i^{y_i}(1-p_i)^{1-y_i},$$ where$\Sigma^{(k)}$denotes the sum over all 0,1 valued$y_i$'s such that $$y_1+\dotsb+y_n=k.$$ As for an approximation, this is not the one you are looking for, but it might help. If ... 1 Let$A=Xw-y$and find the derivative map of the squared norm$L=\|A\|^{2}$:$D_A\|A\|^{2}(H)=\left.\frac{d}{dt}\right|_{0}\|A+tH\|^{2}=\left.\frac{d}{dt}\right|_{0}\langle A+tH,A+tH\rangle=2\langle A,H\rangle$Use the chain rule$D_{w}(L\circ A)=D_{A}L\circ D_{w}A$as follows,$D_w\|A(w)\|^{2}(h)=\left.\frac{d}{dt}\right|_{0}\|A(w+th)\|^{2}=2\langle ...

1

Note that the $\{\varepsilon_t\}$ are independent so $$\operatorname{Cov}(\varepsilon_s,\varepsilon_t) = \begin{cases} \sigma^2_\varepsilon,& s=t\\ 0,& s\ne t.\end{cases}$$ Hence \begin{align} \sigma(0) &= \operatorname{Var}(x_t)\\ &= \operatorname{Var}(\varepsilon_t + m_1\varepsilon_{t-1}+m_2\varepsilon_{t-2})\\ &= \sigma^2_\varepsilon(1 ...

1

Your $\sigma(1)=cov(x_{t},x_{t+1})$ should be $cov(m_1\varepsilon _{t-1} + m_{2} \varepsilon _{t-2} + \varepsilon_{t}, m_1\varepsilon _{t} + m_{2} \varepsilon _{t-1} + \varepsilon_{t+1})$ which can be expanded to $$cov(m_1\varepsilon _{t-1},m_1\varepsilon _{t}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon _{t-1}) + cov(m_1\varepsilon _{t-1},m_2\varepsilon ... 1 The answer to your first question is yes: if c \ne 0 and c \in \mathbb R, then cX \sim \operatorname{Normal}(c\mu, |c|\sigma) if X is normal with mean \mu and standard deviation \sigma. This is because the normal distribution belongs to a location-scale family: its PDF is$$f_X(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-(x-\mu)^2/(2\sigma^2)}, \quad ...

1

There are three types of guests at the party: the guilty one, $X$ innocent matches, $39-X$ innocent non-matches.   The guilty one is (considered) certainly a match; and as for the rest we assume $X$ has a Binomial Probability Distribution. $$X\sim\mathcal{Bin}(39, 0.08)$$ Then we have: \begin{align} \mathsf P(G\mid E) & = \dfrac{\mathsf P(G\cap ... 1 The binomial distribution isn’t really appropriate here. We’re given that exactly one of the guests is guilty, so the events “guest X is guilty” and “guest Y is guilty” are mutually exclusive, which is about as far as you can be from independence. (Remember that the binomial distribution applies to a set of independent Bernoulli trials.) The key to the ... 1 The notation is used to say that a random variable follows a specific distribution, in this case the Chi distribution with parameter (n-1). 1 Your function F is defined on a disc of radius one, which is a compact set. Therefore, the maximums and minimums are either on the boundary of the disc, or strictly inside the disc. Case 1: suppose they are on the boundary. In this case, rewrite your problem in polar coordinates as follows: F(r,t)=2r^2\cos^2(t)-3r^2\sin^2(t)-2r\cos(t), \quad ...

1

You say you take the average of all the averages, but I notice that you have a sample count column. Are these averages over different sample sizes? If so, then you would probably want a weighted average for your aggregate average: $$\text{Aggregate Average} =\frac{\sum_i (\text{sample size})_i(\text{average})_i}{\sum_i (\text{sample size})_i}$$ But without ...

1

I would not guess that a stemplot is the best way to visualize these data, and so I would not recommend one way of setting up the stems as better than another. Particularly so, because the data are strongly skewed to the right, spanning a couple of orders of magnitude. Nevertheless, stem plots can be made: Here is the default stemplot of these 32 ...

1

a) seems to be correct to me. b) Your reasoning isn't clear. Ignore the color of the dice and just label them 1 through six. The first die can be paired to any of five other die. The third die can be paired with any of the three other die and the fifth and sixth die must be paired. This is 15 possible pairings. The first pair can go in any of the three ...

1

No, in general, if $X$ follows a normal distribution, then in shorthand it is written as $$X\sim N(\mu,\sigma^2).$$ Thus, for example, the variance of $X_1$ is 3 not $3^2$. Or as you have written it $$\sigma_1^2 = 3$$ not $\sigma_1 = 3$.

1

Without loss of generality, let's assume $\theta = 0$ and $\sigma = 1$ for simplicity (for the general case, consider the transformation $Z = \frac{X - \theta}{\sigma}$). Rigorously, we have to show $E[|g(X)X|] < \infty$ at beginning. Indeed, denote $\frac{1}{\sqrt{2\pi}}$ by $c$, it follows that \begin{align} & E[|g(X)X|] \\ = & ...

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