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1

Suppose: $A_i$ and $B_i$ for each $i = 1, 2, \ldots$ represent times for A and B to complete their $i^{\rm th}$ race, respectively. Once the $i^{\rm th}$ race is complete, the $(i+1)^{\rm th}$ race is begun immediately, without regard to the status of the other racer (i.e., the two processes are independent). Let $N_A(t)$ and $N_B(t)$ represent the total ...


1

Denote by $F$ the event $B>A$ (B wins before A ) since the events are independent and denote $S$ the event $A>B$ $$\Bbb{P}(B>A, n \text{ times}) = \Bbb{P} (FF \ldots FS) = \bigg(\frac{\alpha}{\alpha + \beta}\bigg)^n \frac{\beta}{\alpha + \beta} $$ since $\Bbb{P}(A>B) = P(A \geq B) = 1 - P(B>A)$


0

Since $\log(x)\lt x-1$, for $x\gt0$ and $x\ne1$ $$ \mathbb{E}(\log(X))\lt\mathbb{E}(X-1)=0 $$ Thus, $$ \begin{align} \prod_{k=1}^n x_k &=\exp\!\left(\sum\limits_{k=1}^n\log(x_k)\right)\\ &\sim\exp\!\left(n\,\mathbb{E}(\log(X))\right)\\[9pt] &\to0 \end{align} $$


2

First write, $$ \begin{align} \lim_{n \to \infty} \prod_{i=1}^{n} X_i &= \exp \left ( \lim_{n \to \infty} n \frac{1}{n} \sum_{i=1}^{n} \log(X_i) \right ) . \end{align} $$ Now by the SLLN we have $n^{-1} \sum_{i=1}^{n} \log(X_i) \stackrel{\text{a.s.}}{\to} \text{E}[\log(X_1)]$. But Jensen's inequality (along with strict concavity of the log function ...


1

Given that $N_m=k$, where $k\ge 1$, and $0\lt x\lt 1$, we have $$\Pr(M_m\le x)=x^k.$$ Now we need to decide what $\Pr(M_m\le x)$ is if $N_m=0$. There is no obvious definition. With not much conviction, we call this probability $1$. Then $$\Pr(M_m\le x)=\sum_{k=0}^\infty e^{-m} \frac{m^k}{k!}x^k.$$ We recognize the sum as $e^{-m}e^{mx}$.


0

I'm very interested in this distribution as well, unfortunately I believe TenaliRaman's answer is only an approximation. Perhaps it only holds when the distribution is sufficiently concentrated to be approximately Gaussian..? I tried the following in R: require(gtools) library(MASS) n <- 7000 C <- c(0,1,0.2,0.5) alpha <- rep(0.2,length(C)) x ...


1

First, $k=1,\dots,n$ (otherwise, the total prob. does not sum up to 1). Then $$\varphi_{X_n}(t)=\mathbb{E}[e^{\mathrm{i}tX_n}]= \sum_{k=1}^{n}e^{\mathrm{i}t\frac{k}{n}}\cdot\frac{1} {n}= \frac{\exp\left(\mathrm{i}t\cdot\frac{n+1}{n}\right)-\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)}{\left(\exp\left(\mathrm{i}t\cdot\frac{1}{n}\right)-1\right)\cdot n} ...


1

well if exponentiation preserves a property then it will continue to hold. For example, $f$ holder cts implies that $e^f$ is so yes that property goes over.


1

You have a minor issue with the expression for $g_Y$: $$g_Y(t)=\frac{q(qt-1)}{p^2+qt-1}$$ Then $$P\{Y=k\}=\frac{g_Y^{(k)}(0)}{k!}=(-1)^{k+1}\frac{(q-1)^2}{(q-2)^{k+1}} \text{, }k=0,1,\dots$$


0

Hmm. Writing integrals as integrals: Say $\epsilon>0$. You need to show there exists $\delta>0$ so $P(A)<\delta$ implies $\int_A|X_n|\,dP<\epsilon$. You know that $\int f(|X_n|)\,dP\le K$ for some fixed $K$. Hmm. Ok. Say $L>0$ is a large number. Choose $M$ so $$\frac{f(t)}t>L\quad(t>M).$$ Now write $A=B\cup C$, where $B=\{t\in ...


1

If you have $\lambda\rightarrow \infty$ then $$\varphi_{Y/\sqrt{\lambda}}(t)=\varphi_{Y}\left(\frac{t}{\sqrt{\lambda}}\right)=e^{-\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}\rightarrow e^{-\frac{t^2}{2}}$$ because $$\left(1-\frac{\lambda(1-\exp\{-\frac{t^2}{2\lambda}\})}{\lambda}\right)^\lambda=e^{-\frac{t^2}{2}}$$


2

\begin{align} \operatorname{E}(e^{itY}\mid X=x) & = \operatorname{E}(e^{it\sqrt{x} \, Z})\quad\text{where }Z\sim N(0,1), \\[10pt] & = \varphi_Z(t\sqrt x) = \exp \left( \frac{-1}2 t^2 x \right). \end{align} \begin{align} \operatorname{E}(e^{itY}) & = \operatorname{E} \left( \operatorname{E}(e^{itY}\mid X) \right) = \operatorname{E}\left( ...


0

The joint density is $$ \text{constant}\cdot \exp\left( \frac{-1} 2 (\mathbf x-\mu)'\Sigma^{-1} (\mathbf x-\mu) \right) $$ where $\mathbf x=\begin{bmatrix} x \\ y \end{bmatrix}$ and $\mu =\begin{bmatrix} \mu_1 \\ \mu_2 \end{bmatrix}$ is the expected value. We have $$ \Sigma^{-1} = \frac 1 {\det\Sigma} \begin{bmatrix} \sigma_Y^2 & -\rho\sigma_{X,Y} \\ ...


0

Intro statistics focuses on the PDF as the description of the population, but in fact it is the CDF (cumulative density function) that gives you a functional understanding of the population, as points on the CDF denote probabilities over a relevant range of measures. If you look at all stats from this perspective, then the PDF is just the description of ...


2

Let $\bar X_n=n^{-1}S_n$. By the CLT you have $$\sqrt{n}(\bar X_n-\mu)\rightarrow N(0,\sigma^2)$$ Now, apply the mean value theorem to get $$g(\bar X_n)-g(\mu)=g'(\bar X_n')(\bar X_n-\mu)$$ where $\bar X_n'$ lies between $\bar X_n$ and $\mu$ so that $g'(\bar X_n')\rightarrow^p g'(\mu)$ (by the WLLN and the continuous mapping theorem) and ...


4

Write $$\frac{S_n - n\mu}{\sqrt{S_n}} = \frac{S_n - n\mu}{\sqrt{n}\sigma}\cdot \frac{\sqrt{n}\sigma}{\sqrt{S_n}}.$$ where $\sigma^2 = \text{Var}(X_1)$. By CLT, $\frac{S_n - n\mu}{\sqrt{n}\sigma} \Rightarrow N(0, 1)$. By WLLN, $\frac{S_n}{n} \to \mu$ in probability. The result follows from the Slutsky's theorem. Clearly, $b^2 = \frac{\sigma^2}{\mu}$.


0

You can use a Snell envelope. Given $\tau$, we can define $U$ recursively with: $U_\tau = V_\tau = 0, \text{and} \\ U_t = V_t \lor E \left[U_{t+1} \mid \mathbb{F}_t \right] \text{for} t = \tau-1, ..., 0$ So, $U$ is a supermartingale which dominates $V$, which means that: $E \left[ \sum_{n=1}^\tau V_n \right] \leq E \left[ \sum_{n=1}^\tau U_n \right] = ...


2

A function $F:\mathbb{R}\to\mathbb{R}$ is a (cumulative) distribution function precisely if (1) $F$ is non-decreasing; (2) $F$ is continuous from the right, that is, $\lim_{x\to a^+}F(x)=F(a)$ for all $a$; (3) $\lim_{x\to-\infty} F(x)=0$; (4) $\lim_{x\to\infty} F(x)=1$. It is not difficult to see that your $F$ has the above properties. Remark: In this ...


1

\begin{align} & \Pr(Y_n \le x) = 1-\Pr(Y_n >x) = 1-\Pr(X_1>x\ \&\ \cdots\ \&\ X_n>x) \\[6pt] = {} & 1 - \Big(\Pr(X_1>x)\Big)^n = 1-(e^{a-x})^n = 1 - e^{n(a-x)}. \\ {} \end{align} It is incorrect to say that $Y_n\le x$ if and only if $\Big( X_1 \le x\ \&\ \cdots\ \&\ X_n\le x \Big)$, but it is correct to say that ...


1

Note that $Y \sim {\mathcal N}(0, \sigma_y^2)$ where $\sigma_y^2 = \sigma_x^2 + \sigma_z^2$ and $\text{Cov}(X,Y) = \sigma_x^2$. Thus $X = \dfrac{\sigma_x^2}{\sigma_y^2} Y + W$ where $W$ is independent of $Y$ and $W \sim {\mathcal N}(0, \sigma_x^2 \sigma_z^2/\sigma_y^2)$. You have $E[X|Y] = \dfrac{\sigma_x^2}{\sigma_y^2} Y$, so $X - E[X|Y] = W$. The rest ...


1

One can show the convergence in $\mathbb L^1$ of $(X_n^tX_n^s)$ to $X^tX^s$ by the estimates $$\mathbb E|X_n^tX_n^s -X^tX^s|\leqslant \mathbb E|(X_n^t-X^t)X_n^s |+\mathbb E|X^t(X_n^s -X^s)|\leqslant \lVert X_n^s\rVert_2\lVert X_n^t-X^t\rVert_2+\lVert X^t\rVert_2\lVert X_n^s-X^s\rVert_2$$ and boundedness of the sequence $\left( \lVert ...


1

Presumably $X$ and $Y$ are jointly normal. Then $$E[X|Y] = \mu_x + \dfrac{\rho \sigma_x}{\sigma_y} (Y - \mu_y)$$ and then $$ E[X e^Y] = E[E[X|Y] e^Y] = \ldots $$ Now use the moment generating function for $Y$: $$E[e^{tY}] = \exp(t \mu_y + t^2 \sigma_y^2/2)$$ and its derivative with respect to $t$, which is $E[Y e^{tY}]$.


1

I am going to write a demonstration that the process is stationary, but I am aware this is not what the original poster asked since it is not based on calculating the characteristic function. This post is in response to a comment I made. Strict sense. It suffices to show that $$\mathbb{P}(z\sin(\omega t+\theta)\le\xi)$$ is independent of $t$. But this is ...


0

Hint: $X+Y = \max\{X,Y\} + \min\{X,Y\}$, so $$\mathbb E[\max\{X,Y\}] = \mathbb E[X]+\mathbb E[Y] - \mathbb E[\min\{X,Y\}], $$ and $\min\{X,Y\}>t$ if and only if $X>t$ and $Y>t$.


1

For $i=1,2,\ldots,\;$ let, \begin{eqnarray*} D_i &=& \text{#Deaths in accident $i$} \\ C_i &=& \text{#Casualties in accident $i$} \\ X_i &=& D_i+C_i. \end{eqnarray*} So we want to find $E(X_i)$ and $Var(X_i)$ to plug into your formulas (with $X=X_i$). \begin{eqnarray*} E(X_i) &=& E(D_i+C_i) \\ &=& E(D_i)+E(C_i) \\ ...


3

A slightly different approach using the law of total probability: \begin{align*} E(X) ={} & E(X\mid(\text{$(i+1)$-th toss is same as $i$-th})\cdot P(\text{$(i+1)$-th is same}) \\ &{} + E(X\mid(\text{$(i+1)$-th toss is different})\cdot P(\text{$(i+1)$-th is different}) \\ = {} & (2)\cdot\frac{1}{6} + (E(X) + 1) \cdot \frac{5}{6} \end{align*} Then ...


4

It's a simple calculation. First, $$ \mathbb{E}[(X-\mathbb{E}[X])^2] = \mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]]. $$ Linearity of expectation implies $$ \mathbb{E}[X^2 + \mathbb{E}[X]^2 - 2X\mathbb{E}[X]] = \mathbb{E}[X^2] + \mathbb{E}[X]^2 - 2\mathbb{E}[X]^2 = \mathbb{E}[X^2] - \mathbb{E}[X]^2, $$ using the fact that $\mathbb{E}[X]$ is a constant. ...


2

The probability of one dice throw to be equal to the previous is constant 1/6, but one previous throw is required to "get started". We would expect to have won once after 6 such tries, but one extra throw is required to get "started". If you want to calculate it using analysis: https://en.wikipedia.org/wiki/Bernoulli_distribution ...


2

From the second roll onwards, the probability that the roll is the same as the previous one is $1/6$. Therefore the number of rolls until two sides show up in a row is distributed $1+\mathrm{Geom}(1/6)$ (one plus a geometric random variable with success probability $1/6$), whose expectation is $1+6 = 7$.


3

Let $Y$ denote the number of rolls that are still needed after the first roll (so $X=1+Y$). Then $\mathbb{E}\left(Y\mid Y>1\right)=1+\mathbb{E}Y$ (do you understand why?) so that $\mathbb{E}Y=\mathbb{E}\left(Y\mid Y=1\right)P\left(Y=1\right)+\mathbb{E}\left(Y\mid Y>1\right)P\left(Y>1\right)=1+\frac{5}{6}\mathbb{E}Y$ We conclude that ...


2

Christian Blatter commented that $E(X)$ simplifies to $\frac{1}{2n-1}\binom{k}{2}$, and suggested there might be an intuitive explanation. Suppose that we have chosen $k$ shoes and lined them up in a row. Let $t=\binom{k}{2}$ be the number of $2$-shoe subsets. For $i=1$ to $t$, let $Y_i=1$ if the $i$-th subset is a matching pair, and let $Y_i=0$ otherwise. ...


3

Let $X_i$ be the indicator random variable for the event that the $i$th pair is picked (i.e. $X_i$ is $1$ if pair $i$ is picked and $0$ if it isn't picked). Then $X = X_1 + X_2 + \ldots + X_n$. Since expectation is linear, we also have $\mathbb{E}(X) = \mathbb{E}(X_1) + \ldots + \mathbb{E}(X_n)$, so we need only find $\mathbb{E}(X_i)$. We have $$ ...


1

Both cars has to be finished before time 2. Thus you have to calculate $P(X+Y<2)$. $$P(X+Y<2)=\int_0^2 \int_0^{2-y}e^{-(x+y)} \, dx \, dy$$ A has to be finished before B. Thus A has time from $0$ to $2-y$. And the finishing time of B has to always greater than the finishing time of A: $y>2-x$ X can vary from 0 to 2. Therefore Y can vary from 0 ...


1

I hope you can do the first part without integration. First A has to take until least t which has probability $e^{-t}$. Then at that time the probability B finishes first is 1/2 by symmetry. The probability B finishes first is the product $(1/2)e^{-t}$. The key thing to remember about the exponential distribution which may be counterintuitive is that it ...


1

Fix $\epsilon>0$. Siince $a_n\rightarrow 0$ and $b_n\rightarrow 0$ there exists $n_0$ s.t. $|a_n|\vee |b_n|<\epsilon$ for all $n\ge n_0$ so that $$P\{|X_n|>\epsilon\}\le P\{|X_n|>|a_n|\vee|b_n|\}\rightarrow 0 \text{ as } n\rightarrow \infty$$ Hence, $X_n\rightarrow 0$ in probability (and in distribution).


3

If $a>0$ then: $$F_Y(x)=P(aX+b\leq x)=P\left(X\leq\frac{x-b}{a}\right)=F_X\left(\frac{x-b}{a}\right)$$ Take the derivative on both sides to find $f_Y$ expressed in $f_X$. Have a look now at the cases $a=0$ (then no density exists) and $a<0$.


2

You allready found that $S$ has binomial distribution with parameters $120$ and $\frac{1}{2}\frac{1}{3}=\frac{1}{6}$. Likewise you can find that $T$ has binomial distribution with parameters $120$ and $\frac{1}{2}\frac{2}{3}=\frac{1}{3}$. That enables you to find an expression for $P\left(T=30\right)$. We have $$P\left(S=s\mid ...


2

$$ \begin{align} Pr(A\leq B) &=\int_{0}^{\infty}f_B(y)Pr(A\leq y)dy\\ &=\int_{0}^{\infty}\lambda_B e^{-\lambda_B y}(1-e^{-\lambda_A y})dy\\ &=\frac{\lambda_A}{\lambda_A+\lambda_B} \end{align} $$


3

Since $A$ and $B$ are independent, their joint density is: $$p(a, b) = \lambda_A\lambda_Be^{-(\lambda_Aa+\lambda_Bb)}$$ In general, if we have a joint density $p(a, b)$ defined on non-negative reals, we have $$P(A < B) = \int_0^\infty \int_0^b p(a, b) da db$$ So \begin{eqnarray*} P(A < B) &=& \int_0^\infty \int_0^b p(a, b) da db \\ ...


1

According to me the problem in your second solution lies in calculation of probability of atleast one accident in a given month. Prob(atleast one) is simply 1 - Prob(None) = 1 - 0.7 = 0.3 We cannot assume that probability of k accidents shall be (0.3)^k because there is no information about the distribution and independence of intra-month accidents. With ...


2

There is an easier approach. Note that $\Phi(x)$ is a continuous increasing function going from $0$ to $1$. Let $Y=\Phi(X)$, so $Y$ is in the interval $(0,1)$. Then $$F(y)=\Pr(Y \le y) =\Pr(\Phi(X) \le y) = y$$ so $f(y)=1$ when $y \in (0,1)$ and $E[Y]=\int_0^1 y \,f(y) \, dy =\frac12$. This works for any continuous distribution.


0

Sketch the graph of $y=cos(x)$ for $-1\leq x\leq 1$. The support of $Y$ will be $(\cos 1; 1]$.   This will be a fold of two halves of the support of $X$.   We map back to these intervals by two semiinversions: $x_1={+}\arccos(y)$ and $x_2={-}\arccos y$. Thus use the change of variable transformation: $$f_Y(y) = ...


0

By linearity of expectation operator, we have: \begin{equation} E\left[\frac{a X_{1} + b X_{2} + c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] = E \left[\frac{a X_{1} + b X_{2}}{e X_{1} + f X_{2}}\right] + E \left[\frac{c X_{3} + d X_{4}}{e X_{1} + f X_{2}}\right] \quad (1) \end{equation} Note that the random variables $cX_3 + dX_4$ and $\frac{1}{eX_1 + ...


0

No, as far as I know this is not the case in general. There is a counterexample for $n=2$, that is related to this example in Wikipedia: $(X_1,X_2)$ are jointly Gaussian with variance $\sigma$ and vanishing covariance (i.e., $X_1$ and $X_2$ are uncorrelated and independent). $Y_1$ is Gaussian with variance $\sigma^2$. $Y_2=WY_1$, where $W=1$ w.p. 0.5 and ...


0

You are probably using an unbiased estimator of the variance, so the $\sigma^2$ you're calculating is actually 6/5 of what you would expect, as you pointed out yourself. Take a look at Wikipedia here and here.


1

My Trial $$\begin{align} F_Y(y) & = P(Y\leq y) \\[1ex] & = P(X^2\leq y) \\[1ex] & = P(X\leq\surd y)\\[1ex] & = F_X(\surd y) \\[1ex] & = \int_0^{\surd y} t \;\mathrm d t & \bigstar \\[1ex] & = 0.5 y\\[2ex] \therefore f_Y(y) & =0.5 \end{align}$$ $\bigstar$ Your error is here. $f_X(t) = 1$ for all $t\in[0;1]$, and $F_X(t) = ...


5

The method is fine in principle, but the wrong density function for $X$ was used..\ As in your argument, we have, for $0\lt y\lt 1$, $$F_Y(y)=\Pr(X\le \sqrt{Y}).$$ However, if you really want to use an integral, $$F_X(\sqrt{y})=\int_0^{\sqrt{y}} 1\cdot dt=\sqrt{y}.$$ This is because the density function of $X$ on $(0,1)$ is $1$. Differentiating, we find ...


0

HINT To compute the characteristic function we can use the so called tower rule: $$E\left[e^{j\times v \times ξ(t)}\right]=E\left[E\left[e^{j\times v\times Z \times \sin(ωt + θ)} \mid Z\right]\right].$$ Since $Z$ and $\theta$ are independent we can tell that $$E\left[e^{j\times v\times Z\times \sin(ωt + θ)}\mid Z=z\right]=E\left[e^{j\times v \times ...


0

Break it down. For Player $A$ to even have a chance of being on a team with $k$ players (for $k<n$), you need $X=k$ or $X=n-k$. These two happen with probability $1/(n-1)$. Start with $X=k$. We need player $A$ to land in the first team, which happens with probability $\binom{n-1}{k-1}/\binom{n}{k}=\frac{k}{n}$: we set aside Player A and pick the rest of ...



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