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0

As there is a law of total expectation, so there exists a law of total variance. We will call the random variable representing the total number of widgets ordered in a month $W$. Note that $$W = Y_1 + Y_2 + \ldots, Y_X,$$ because if we observe $X$ orders in a month, that means we observed orders $Y_1, Y_2, \ldots, Y_X$ of widgets in each order. So, the ...


1

Hint: for calculating $p(x = 2)$, you don't need to work out $p(AC), p(BC), p(DC)$ separately, you just need to work out $p((\text{not-}C)C)$, which is $\frac{8}{10}\frac{2}{9}$. Can you see how to generalise this? Hint for part 2: work out the conditional probability $p(\text{the fish of type D gets taken out}|x=r)$, then use the law of total probability ...


2

One rejects the null hypothesis when the P-value is small. A common criterion is to reject if the P-values is less than 0.05. In a Kolmogorov-Smirnov test, the D-statistic measures the maximum diagonal distance between the empirical cumulative distribution functions (ECDFs) of the two samples. (Everything is re-scaled so the ECDF fits inside the unit ...


0

Isn't a good approach to calculate the average wait time if B arrives at 2pm (50% probability) and the average wait time if B doesn't arrive at 2pm (50% probability), then average these two average times out, in 50% - 50% proportion. If B arrives exactly at 2pm, then B's average wait will be 30 minutes. If B doesn't arrive exactly at 2pm, I believe the ...


1

It is easier to calculate the expectation directly than it is to write down the distribution. This is usually the case. Since you haven't said that much about what you've tried yet, let me start with a hint. Try to calculate the expected time to hit $\{ 0,n \}$ starting from $x$ with $0<x<n$. Then send $n \to \infty$. You can set up a system of ...


1

The probability of $X=x$ tests needed is the sum over the number of students $n$ who fail the original test of: the probability of $100-n$ passes and $n$ fails in the $100$ original test multiplied by the probability of $n-1$ passes and $x-100-n$ fails in any order in the first $x-101$ retests multiplied by the probability the final retest is a pass so ...


2

Try these formulas: \begin{eqnarray*} B1 &:& \quad\text{=FLOOR(RAND()*(A1+1),1)} \\ C1 &:& \quad\text{=FLOOR(RAND()*(A1+1),1)} \\ D1 &:& \quad\text{=MIN(B1,C1)} \\ E1 &:& \quad\text{=MAX(B1,C1)-D1} \\ F1 &:& \quad\text{=A1-E1-D1} \\ \end{eqnarray*} Columns $D1,E1,F1$ are the three numbers required. $A1,B1$ are two ...


2

Do you know what a $\sigma$-algebra is? It is a collection of subsets of the sample space $\Omega$. In particular, it is a collection of subsets that a) contains $\Omega$ as an element, b) is closed under set complement, and c) is closed under countable unions. The purpose of a $\sigma$-algebra is to act as the domain of a measure $P$. So the measure ...


1

A random varible is defined on a probability space which consists of a sample space $\Omega$, a $\sigma$-algebra $\mathcal{F}$ on $\Omega$ and a measure $P:\mathcal{F}\rightarrow [0,1]$. In this case, $X$ is a random variable if for ever $x\in\mathbb{R}$, $[X\le x]\in \mathcal{F}$. To wit, $X$ is a measurable function on the measure space $(\Omega, ...


2

If $X_i \sim \operatorname{Exponential}(\lambda)$ are iid such that $$f_{X_i}(x) = \lambda e^{-\lambda x}, \quad x > 0,$$ then $T = \sum_{i=1}^n X_i \sim \operatorname{Gamma}(n,\lambda)$ with $$f_T(x) = \frac{\lambda^n x^{n-1} e^{-\lambda x}}{\Gamma(n)}, \quad x > 0.$$ Then it is easy to see that $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} ...


2

It may help to use the fact that $$\mathbb{E}\left[X\right] = \int_0^{\infty} \mathbb{P}(X \geq x) dx$$ for positive continuous variables so that $$\mathbb{E}\left[\frac{n}{X_1+\ldots+X_n}\right] = \int_0^{\infty} \mathbb{P}\left(\frac{n}{X_1+\ldots+X_n} \geq x\right) dx = \int_0^{\infty} \mathbb{P}\left(X_1+\ldots+X_n \leq \frac{n}{x}\right) dx$$ Now use ...


1

the first step is to invert the covariance matrix so that you can express the joint distribution function as $$ f(x_1,f_2) = \frac{1}{N} e^{-(A_{11} x_1^2 + A_{12}x_1 x_2+ A_{22}x_2^2)} $$ where $N$ is a normalization constant. Then by integrating from $x_1 = 0$ to infinity you find the conditional distribution of $x_2$ given that $x_1 > 0$ will be ...


0

Let $\Omega_X$ be the set of full measure on which $X_n \rightarrow X$. Define $\Omega_Y$ likewise. Note that $\Omega_0:=\Omega_X \cap \Omega_Y$ has full measure as well. On $\Omega_0$, $X_n \rightarrow X$ AND $X_n \rightarrow Y$. Hence $X=Y$ a.s due to uniqueness of limit.


1

Suppose that $X_n\to X$ almost surely as $n\to\infty$ and $X_n\to Y$ almost surely as $n\to\infty$. Then there exists $\Omega'\subset\Omega$ such that $\Pr(\Omega')=1$ and for each $\omega\in\Omega'$ $$ |X_n(\omega)-X(\omega)|\to0 $$ as $n\to\infty$. Similarly, there exists $\Omega''\subset\Omega$ such that $\Pr(\Omega'')=1$ and for each $\omega\in\Omega''$ ...


1

Here is an answer with the simplest notation I can manage. Maybe you can match ideas here with the content of the previous Answer by @r.e.s. You want to estimate the probability $p_1$ that $X = 1$ based on a sample of $n$ independent observations from the distribution of $X$. You count $Y_n$, the number of instances among $n$ in which $X = 1.$ Then $Y_n ...


1

HINT: Proportions are intuitive estimators of probabilities; i.e., to estimate $P(X \in A)$ given i.i.d. observations $X_1,...,X_n$ of $X$, consider the proportion of the $n$ observations that are in $A$: $$\hat{P}(X\in A) = \frac{1_{X_1\in A} + 1_{X_2\in A} +\ ... +\ 1_{X_n\in A}}{n}, $$ where $$1_E = \begin{cases} 1, & \text{if E occurs} \\ 0, ...


3

Given that $Y$ is Bernoulli distributed, it will have probability mass function $P(Y=0)=(1-p)$ and $P(Y=1)=p$. As Frank has helpfully commented, we can condition on $Y$. When $Y=0$, we have $P(\frac{X}{Y-X}<0|Y=0)=P(-X<0|Y=0)=P(X>0|Y=0)$ When $Y=1$, we $P(\frac{X}{Y-X}<0|Y=1)=P(\frac{X}{1-X}<0|Y=1)=P(X>1|Y=1)$ Thus, we have the ...


0

We assume that the PDF of $v(t)$, namely $f(x, t)$, always has a finite first moment (that is, the mean). Let us consider the behavior of the position from $t = 0$ to $t = \Delta t$, where $\Delta t$ is chosen small enough that the variation in the first moment is as small as we like. (That we can do this is established by the continuity of $f(x, t)$.) As ...


0

When you assert that you can equivalently formulate a set of $n$ Bernoulli variates as a finite sequence of length $n$ such that the probability of each possible sequence is equal, you have imposed conditions that do imply the independence of the $n$ Bernoulli variates. On the other hand, suppose for example you take a space of sequences such that the ...


0

Consider the $2^n$ possible sequences of heads and tails. If by definition these outcomes are equally likely then the model coincides with the one based on distinct events and assuming independence. Saying that the possible sequences are equally likely is equivalent to the assumption of independence. Most importantly: both assumptions are arbitrary. You ...


1

How is the time till bankruptcy distributed? This is an application of the Hitting Time Theorem (see, e.g. here (Theorem 1) or pg. 79 of Grimmett and Stirzaker). $$P(\text{Ruined at game $n$ starting with $\$x$}) = \dfrac{x}{n}\binom{n}{(n-x)/2}p^{(n-x)/2}q^{(n+x)/2}.$$ Is the expected time till bankruptcy $= \infty$? Yes, if $p\geq q$. ...


0

Covering is a standard technique to obtain bounds on the supremum of random processes. In Lemma 10 the author is essentially showing that the map $S$ is isometric restricted to the subspace. Isometry is ensured by the fact that $S$ is a JL transform. JL transform applies to a finite set of points however if you cover the subspace (rather the unit sphere over ...


0

Fortunately, the scenarios are in increasing order of profit for both investments. This simplifies matters. If the profits under scenario $i$ are denoted by $p_i$, then the PDF (or more accurately, the probability mass function, or PMF) of the profit is simply given by the scenario probability distributions: $$ f(p_i) = P(\mbox{scenario } i) $$ For ...


3

Very good questions. If $X$ and $Y$ have the same probability distribution, than for every $z$, $$ \Pr\left(X \leqslant z\right) = \Pr\left(Y \leqslant z\right) $$ Then $X$ and $Y$ are said to be identically distributed. Note that this does not imply independence. A classic example would be binormally distributed vector of $(X,Y) \sim ...


1

You want to find $c$ such that $$-1\le \dfrac{\sqrt{(Y-2/c)^2-4}}{2} \le 1$$ (and in particular is real) for $0 \le Y \le 1$. Equivalently, $$ 4 \le (Y - 2/c)^2 \le 8$$ for $0 \le Y \le 1$. So either $2 \le Y - 2/c \le 2 \sqrt{2}$ in that interval, or $-2\sqrt{2} \le Y - 2/c \le -2$. Now unfortunately, $2\sqrt{2} - 2 < 1$, so this can't ever be true ...


1

If $0 < Y < 1$, then $0^2 < Y^2 < 1^2$, so it follows that $0 < Z < 1$. More generally, if $a < Y < b$, then we must address three cases: $$\begin{align} 0 \le a < b, \\ a < 0 < b, \\ a < b \le 0. \end{align}$$ In the first case, it should be clear that $$a^2 < Z < b^2.$$ In the second, because $0 \in (a,b)$, we ...


4

Observe that for each $n$, $\{X<\infty\} = \{\sup_{k\ge n}X_k <\infty\}$.


-1

What is the probability that you try to check your email during the first $x$ minutes of the $(x+y)$-minute cycle? If you do that, then your waiting time is $0$. What is the probability that you try to check your email when more than $w$ minutes, but fewer than $y$ minutes, remain of the $(x+y)$-minute cycle, where $0<w<y$? In that case, you need to ...


2

Your model is different. If he buys a ticket with the winning number from the winning lottery, probability that it will appear next time, is $10^{-5}$. Now you have easy to solve inequality $$ \left(1-10^{-5}\right)^n\leq 0.5. $$ (And the answer to your second question is: yes. If choosing the winning ticket is really random, it is meaningless, which one ...


0

$P($at least one success$) + P($no success$)=1 \implies P($at least one success$)=1-(0.88)^n$. I think you can proceed from here.


0

Hint: $P(X \geq 1)=1-P(X=0)=1-0.88^n\geq 0.95$


2

Oops, you made a mistake. $P(X\geq1)=1-P(X=0)$. You have it as $P(X\geq1)=1-P(X=0)-P(X=1)$


1

For the last series, since the second is convergent it suffices to check that the series $\sum_{n\geqslant 1}b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]$ is convergent. To this aim, we start writing that $$b_n^2\mathbb E\left[X_1^2\mathbf 1\{|b_nX_1|\leqslant 1 \}\right]\leqslant \int_0^1 t\mathbb P\{|b_nX_1|\geqslant t\}\mathrm dt.$$ ...


2

So we have $P(X>Y) = P(X-Y>0) = P ((X,Y) \in C)$ Where $C = \{(x,y):x-y>0\}$ But we know how to find this, as this is the same as simply evaluating f(x,y) and integrating this over the region C. $P(X>Y) = \iint_C f(x,y) dxdy$ .


1

You must calculate $$\int_{-\infty}^{+\infty}\left[\int_y^{+\infty}f(x,y)dx\right]dy$$ or $$\int_{-\infty}^{+\infty}\left[\int_{-\infty}^xf(x,y)dy\right]dx$$ which yeld the same result.


2

What you want is $|cY|=|c|\cdot |Y| \le 1$; knowing that $|Y|\le 1$ and that the boundary can be reached, it is necessary and sufficient to have $|c|\le 1$. So your second answer, $[-1,1]$, is correct.


0

For the last part you just need to set the limits of integration correctly. The limits for $X$ ($0$ to $1$) are all the $X$ values for which $X+Y\lt 1$ is possible. Then, for $Y$, we treat $X$ as a constant between $0$ and $1$ and the limits of integration for $Y$ should cover all possible values for $Y$ that satisfy $X+Y\lt 1$. So, \begin{eqnarray*} ...


1

So if I understand right, if $X-\mathbb EX$ is $\sigma^2$-sub-Gaussian, then $$\forall t\in\mathbb R:\mathbb Ee^{t(X-\mathbb E X)}\le e^{t^2\sigma^2/2}\tag1$$ Taking the logarithm of both sides of (1), we have an expression in terms of the so-called cumulant-generating function of $X$: $$\forall t\in\mathbb R:\log\mathbb Ee^{tX}-\mathbb EtX\le ...


1

Since the sequence $(X_n)_{n\geqslant 1}$ is independent, an application of the Borel-Cantelli lemma shows that $X_n\to 0$ almost surely is equivalent to the convergence of the series $\sum_{n\geqslant 1}\mathbb P(|X_n|>\varepsilon)$ for each fixed $\varepsilon$. Thus, $X_n\to 0$ almost surely is and only if $\sum_{n=1}^\infty\mathbb ...


0

In regards to question $(2)$: We have that $p_{i,j} > 0\ \forall i,j$, consider $$p_{i,j}^{(2)} = \sum_{k}p^{}_{i,k}p_{k,j} > 0,\text{ since }p > 0 \ \forall i,j$$ With an induction step, we can show that $p^{(n)}_{i,j} > 0 \ \forall i,j,n$. Let us assume that $p^{(n)}_{i,j} > 0\ \forall i,j$ for some $n$, we will show that $p^{(n+1)}_{i,j} ...


1

(a) Your answer is right until the last step. Because $N(t)$ has a Poisson distribution with parameter $\lambda t$, it should be: $$P(N(t)=1)\cdot P(N(t)=1)\cdot P(N(t)=1) = (\lambda t \;e^{-\lambda t})^3.$$ (b) Your answer is right. (c) \begin{eqnarray*} E(T_2\mid T_1\lt T_2\lt T_3) &=& \int_{t_2}{t_2 P(T_2=t_2\mid T_1\lt T_2\lt T_3)\;dt_2} \\ ...


1

As you say, the explanation is terse. You correctly understand why you can say $4\theta_1 + 6\theta_2 = 1,$ and the formula $2\theta_1 + 3\theta_2 = \frac12$ appears to be, as you say, just dividing that formula by $2$. But another motivation for writing the formula that way is that the sum of probabilities in a single column of the joint probability ...


2

I think the question is asking whether you can get different probability distributions by choosing different values of $c$. If you can, then $c$ would be a "parameter". However, in this case, there is only one value of $c$ that makes $f$ be a probability distribution at all (because the total probability has to be $1$), so it is not a parameter, but must ...


1

For $N\sim\text{Geometric}(p)$, we wish to compute $\mathbb E[\gamma^N]$ for $\gamma:=\frac{pe^t}{1-qe^t}$. Now, since $\left|p\frac{qe^t}{1-qe^t}\right|$$<1$ for $t<-\log q$, observe that the infinite sum of a geometric series justifies $$\mathbb ...


1

I finally find what the problem is, which is the possibility of $f(r)$ being a probability density function. It depends on what parameters, $\mu$, $\lambda$ and $x_0$, we choose. In the adjusted code, I set $\mu=0.5$, $\lambda=1$ and $x_0=0.945$ to assure $f(r)$ being a feasible density, then set the scale factor $M=2$, which work perfectly. Here is the ...


1

Let $X$ be the event that the packet is transmitted error-free. The probability that none of $n$ bits is transmitted wrong (assuming errors in different bit positions occur independently), is simply $P(X|N=n)=(1-p)^n$. To obtain $P(X)$, we need to add over all $n$ $$\begin{align} P(X)&=\sum_{n=1}^\infty P(X|N=n)P(N=n)\\&=\sum_{n=1}^\infty ...


2

$$\sum_{n\geq k}\frac{1}{n!}=\frac{1}{k!}\left(1+\frac{1}{k+1}+\frac{1}{(k+1)(k+2)}+\ldots\right)\leq \frac{1}{k!}\sum_{j\geq 0}\frac{1}{(k+1)^j}=\frac{k+1}{k\cdot k!}$$ hence we just need to prove that: $$ \left(1+\frac{1}{k}\right)\frac{k^k}{k!}\leq e^k. \tag{1} $$ The inequality triviall holds for $k=1$; moreover, by setting $A_k = ...


1

This is a good observation. The distinction here is that the elements of $L^2$ are not actually functions, but equivalence classes of functions. In this case, the zero element of $L^2$ is $$\{X\in L^2 : \mathbb P(X=0)=1\}. $$ As $\langle X,X\rangle=0$ implies that $\mathbb P(X=0)=1$, positive-definitenes holds.


1

Bruce Trumbo's answer is good, but if you want a more basic approach then note that the probability of a single observation of being between $0.25$ and $0.75$ is $\frac12$, being low is $\frac14$, and being high is $\frac14$. The middle value will be in the central interval if any of the following patterns happen: all three observations central, with ...


0

Let $M(t)= E[e^{tX}]$ Then $$Ent(e^{tX}) = E[e^{tX}tX] - E[e^{tX}]log(E[e^{tX}]) = tM'(t) - M(t)log(M(t)) = Mt^2\frac{d}{dt}\frac{log(M(t))}{t}$$ Given: $$E[e^{t(X-E[X])}] \leq \frac{e^{t^2 \sigma^2}}{2} \implies M(t) \leq e^{\frac{t^2 \sigma^2}{2}+tE[X]} \implies log(M(t)) \leq \frac{t^2 \sigma^2}{2} +tE[X]$$ Thus, for $t \neq 0$ , ...



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