New answers tagged

0

Hint: $P(B) = 0.04$ $P(G) = 0.01$ Where B and G are the events that boy and girl, are respectively taller than $1.8$m. There are $40$% boys and $60$% girls. The probability you are looking for is $$P = \frac{? \cdot 0.01}{? \cdot 0.04 + ? \cdot 0.01}$$


1

Just integrate piecewise: $$\int f(x) \mathrm{d} x = \int_0^2 c \text{d}x + \int_5^{10} 2c \text{d}x = 2c+10c=12c.$$ So take $c= \frac{1}{12}$.


0

It looks like you want to design a quantizer. A good textbook on this is "Vector Quantization and Signal Compression" by Gersho and Grey. One way to design such a mapping is to define a function (called a quantizer) $g(x)$ (which gives the approximated value) and then minimize some average of a loss function $L(x,g(x))$ (i.e. minimize $E[L(X,g(X))]$). ...


2

No. By your definition, $U_1,U_2$ will be both uniform on the unit interval. Then, by Holder's inequality $E[U_1U_2]\leq (E[U_1^p])^{1/p}(E[U_2^{(1-1/p)^{-1}}])^{1-1/p} = \left( \frac{1}{p+1} \right)^{1/p} \left( \frac{1}{\left(1-1/p \right)^{-1} +1} \right)^{1-1/p}$ for all $p\geq 1 $. Now, plot the upper bound, and see that there exists a $p$ such that ...


0

The random variables $U_i = \Phi(X_i)$ are $U(0,1)$ uniformly distributed since $\Phi$ is one-to-one and $$P(U_i \leqslant x ) = P(\Phi(X_i) \leqslant x ) = P(X_i \leqslant \Phi^{-1}(x)) = x.$$ We have $E(U_i) = 1/2$ and $$cov(U_1,U_2) = E(U_1U_2) - 1/4 = \int_{-\infty}^\infty\int_{-\infty}^\infty\Phi(x_1)\Phi(x_2)n(x_1,x_2,0.5) \, dx_1\, dx_2 - 1/4,$$ ...


0

The Gaussian copula (with correlation matrix $R$): $$ C_R(u)=\Phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R). $$ Then, letting $\zeta:=(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n))^{\top}$, $$ c_R(u)=\frac{\phi_n(\Phi^{-1}(u_1),\dots,\Phi^{-1}(u_n);R)}{\phi(\Phi^{-1}(u_1))\cdots \phi(\Phi^{-1}(u_1))} \\=(2\pi)^{-n/2}\lvert R ...


0

Which is not the expectation for a geometric distribution. Why can't a geometric distribution be applied to this problem? The definition I have been given for a geometric random variable is the number of independent trials until the first "success". Where each trial has a probability of p. What would be the correct way to determine the distribution for ...


0

You can setup a system of linear equations and inequalities that characterizes the set of all feasible solutions. Given $P(X=k),$ $k\ge 0$, $P(Y=h),$ $h\ge 0$ and $E[XY]$ this system will be $$ \begin{array}{rcl} \sum\limits_{h\ge0}g_{h,k}&=&P(X=k),\ k\ge 0,\\ \sum\limits_{k\ge0}g_{h.k}&=&P(Y=h),\ h\ge 0,\\ \sum\limits_{k,h\ge ...


1

Andre is right about why this is not a geometric distribution. But actually, the expectation given in your solution is also incorrect. The correct answer is the the question posed is 5.4. The point is that if after guess 9 you have not received any answer of "yes" then you can deduce for sure that the prize is in the remaining unasked-about box.


0

Suppose $N$ is the total number of tickets, $k$ is the number of winning tickets, and $n$ is the number of tickets you buy. In your case $N=100, k=2, n\geq k$. As suggested in the comment above, the distribution of the random variable $X$= # of winning tickets after $n$ draws follows the hypergeometric distribution. Clearly the total number of ways you can ...


1

If the draws are independent, this is correct.


1

Your attempt is almost correct. You have to show that there exists for each $\epsilon>0$ a compact set $K_\epsilon$ such that for all $n$, the probability that $X_n\in K$is at least $1-\epsilon$. Since every compact subset of $\mathbb{R}$ is contained in some bounded closed interval, and since the probability that an exponentially distributed random ...


0

P(win both prizes)$=\frac{number\;of\;ways\;choosing\;n\;tickets\;containing\;both\;tickets }{number\;of\;ways\;choosing\;n\;tickets\;}$ There are $2\choose2$$100-2\choose n-2$ ways to choose the rest (n-2) tickets in order to win. There are $100\choose n$ ways in total to choose n tickets. So, P(win both prizes)=$\frac{{2 \choose 2}{100-2\choose ...


0

There are $n \choose 2$ ways to select the winning tickets from $n$ and ${100 \choose 2}$ overall ways to select the winning tickets, assuming the winners are interchangeable.


0

You have a number regions: $1\le z\le 3$: $$ F_Z(z)=\int_{2-z}^1\int_1^{\frac{z+y}{2}}\frac{x}{8}dxdy=\frac{1}{192}z^3+\frac{1}{64}z^2-\frac{1}{64}z+\frac{5}{192} $$ $3< z\le 5$: $$ F_Z(z)=\int_{-1}^1\int_1^{\frac{z+y}{2}}\frac{x}{8}dxdy=\frac{1}{32}z^2-\frac{11}{96} $$ $5< z\le 7$: $$ ...


0

If $1\le z\le3$ the limits on your double integral are wrong. It is true in the inner integral that $x$ runs from $1$ to $\frac{z+y}2$, but in the outer integral $y$ runs from $2-z$ to $1$. The lower limit for $y$ is $2-z$ because that's the value of $y$ when the line $2x-y=z$ intersects the line $x=1$. (The lower limit for $y$ is not $-1$ unless $z=3$.) ...


1

In general, if $Z$ is a random variable in the probabilistic space $(\Omega,\mathcal{F},P)$, and $X=bZ$then, $$\int_\Omega X^ndP=\int_\Omega b^nZ^ndP=b^n\int_{\Omega}Z^ndP$$. Thus $E[X^n]=b^nE[Z^n]$.


1

It's not necessary to prove that $\lim_{n\to\infty}\mathbb{P}(X_n\leq c)=0$. Instead, let $$A=\{\alpha\in\mathbb{R}:\lim_{n\to\infty}\mathbb{P}(X_n\leq \alpha)=0\} $$ and observe that if $\alpha<c$, then there exists $\gamma>0$ such that $\alpha<c-\gamma<c$, hence $$\mathbb{P}(X_n\leq \alpha)\leq \mathbb{P}(X_n<c-\gamma)$$ for all $n$, so $$ ...


0

First of all you need for the probability density function $f_X(x) \geq 0$ and for it to integrate to $1$. $$\int_{-\infty}^{\infty}f_X(x)\:\text{dx} = \int_{x \in D}{}f_X(x)\:\text{dx} = 1$$ Then for the moments $$E[X^n] = \int_{-\infty}^{\infty}x^n f_X(x)\:\text{dx} = \int_{x \in D}{}x^nf_X(x)\:\text{dx} $$ and since you used the moment-generating ...


2

$\lim_{x \to \infty} 2e^{-2x} = 0$ And we have $$P(X>1) = \int_1^{\infty} 2e^{-2x} = \frac{1}{e^2}$$ You must have missed something when you did the integration Finally $$\int_0^{\infty} 2e^{-2x} = 1$$


0

I saw the question and answer from one of the website about the b) from https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.408149.html Note: The probability of x successes in n trials is: In this case p = .05 & q = .95 a) P(exactly 3 of 20 being defective computers) = 1140(.05)^3(.95)17 = ...


1

Let X_n be the random variable corresponding to the number of +1 steps in n turns. $X_n$ will follow Binomial Distribution with parameters $n$ and $p$. If $k$ is the number of +1 steps. Then, $n-k$ is the number of +2 steps. Then, $k+2(n-k) =t$. On solving, we get $k = 2n - t$. Thus, $P(T_n = t) = P(X_n = 2n-t) = {n \choose 2n-t}p^{2n-t}(1-p)^{t-n}$. Now, ...


1

Note that $\frac{S_i+2}{3}$ is distributed as $B(1,p)$, giving that $\frac{T_n+2n}{3}$ is distributed as $B(n,p)$. Thus: $$\mathbb{E}\left(\frac{T_n+2n}{3}\right)=np, \mathbb{V}\left(\frac{T_n+2n}{3}\right)=np(1-p)$$ from which you can derive: $$\mathbb{E}(T_n)=n(3p-2), \mathbb{V}(T_n)=9np(1-p)$$ Furthermore, knowledge of the binomial distribution gives ...


1

$P(T_n=t)=P(\sum_{i=1}^n S_i=t)$ so this is essentially the $n$-fold convolution of $S_1,S_2,...,S_n$. Not sure if a neat answer exists. $E(T_n)=E(\sum_{i=1}^nS_i)=\sum_{i=1}^nE(S_i)=\sum_{i=1}^n[1\times p-2\times(1-p)]=n(3p-2)$ $Var(T_n)=Var(\sum_{i=1}^nS_i)=\sum_{i=1}^nVar(S_i)=nVar(S_1) $ due to iid $S_i$. Find $Var(S_1)=E(S_1^2)-(E(S_1))^2$. Simplify. ...


0

[Assuming the dice is otherwise fair - that is, if you just rolled $x$, the next roll is discrete uniform on $\{1,2,3,4,5,6\}-x$] Say the rolls are ordered $X_1,X_2,\cdots$ and $x_i\in{1,2,3,4,5,6}$ for all $i$. Then the joint pdf is: $$\mathbb{P}(X_1=x_1, X_2=x_2,\cdots,X_n=x_n)=\frac{1}{6}\left(\frac{1}{5}\right)^{n-1}\prod_{i=1}^{n-1}\mathbb{I}\{x_i\neq ...


1

For ordinary simulation, a box-muller/polar marsaglia/ziggurat algorithm with cholesky decomposition allows you to simulate a bivariate normal very efficiently. However as you said you want to apply the central limit theorem, which seems that you want to simulate a lot of random variables and show the convergence. Assume you already have a random number ...


2

Split it into disjoint events, and add up their probabilities: The probability of $\color\red0$ heads in each part is $\frac{\binom{3}{\color\red0}}{2^3}\cdot\frac{\binom{3}{\color\red0}}{2^3}$ The probability of $\color\red1$ heads in each part is $\frac{\binom{3}{\color\red1}}{2^3}\cdot\frac{\binom{3}{\color\red1}}{2^3}$ The probability of $\color\red2$ ...


0

Hint: Let $X$ be the number of heads in the first three rolls, and $Y$ be the number of rolls in the last three. Then $X$ and $Y$ are independent and follow what kind of distribution? Further, we are asked to calculated $$P(X = Y) = \sum_{k = 0}^3P(X = k, Y=k).$$


0

$X$ can be written as a sum $X_{s} + X_a$, where $X_{s}$ is symmetric and $X_a$ is antisymmetric. But $a X_a a^\text{T} = 0$ for any $a$ and any antisymmetric $X_a$, so WLOG, suppose $X = X_s$. Then your penultimate sentence applies.


0

Well, both are wrong. Take 100 brand new disks, each with an expectancy of 100,000 hours to the first failure. How long does it take until the first one fails? Certainly before 100,000 hours, but you'd need to know at least the variance, and better the distribution. You might have disks that all without exception fail after 99,000 to 101,000 hours, then ...


0

The $1-\alpha$ confidence interval for $\hat p$ is $$\large{\left[ \hat p-z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} , \hat p+z_{(1-\frac{\alpha}{2})} \cdot \sqrt{\frac{\hat p(1-\hat p)}{ n}} \right]}$$ $z_{(1-\frac{\alpha}{2})}$ is the z-value of the standard normal distribution. $1-\alpha$ is the confidence level. In your case ...


1

It is a one-dimensional normal distribution. https://en.wikipedia.org/wiki/Cram%C3%A9r%E2%80%93Wold_theorem This follows from an alternative definition of multi-dimensional Gaussian distribution. EDIT: see the Wikipedia article (for future reference/anyone new to the question) https://en.wikipedia.org/wiki/Multivariate_normal_distribution#Definition


0

Assuming they are real valued: $P(X=Y)=\int_{-\infty}^\infty P(Y=x|X=x) \ P(dx)=\int_{-\infty}^\infty P(Y=x) \ P(dx)=0$, because $P(Y=x)=0$ for any $x\in\mathbb{R}$ since $Y$ has a continuous distribution (pdf). If you meant for the cdf to be continuous, then the answer changes. In general, though $P(X=Y)>0$ is possible, but you have to relax some of ...


4

Compare it to lottery. If you own one ticket, the probability to win is very low. If you own 100 tickets it's much more likely that you win (something). Likewise, the probability that one disk fails is quite low but the probability that any of the 100 disks fails is higher, and therefore the mean time to failure is lower.


3

The mean time to failure of some disk is not the mean time to failure of those disks: it is the mean time before at least one of those 100 disks fails. In other words, the time to failure of some disk $T$ is the minimum of the time to failure of all disks, i.e. $$T = \min_{i \leq i \leq N} \{X_i\}.$$ A standard of model is that aging has no memory ...


8

I think that's a misunderstanding. The passage you quote is a bit informally phrased. By "the mean time to failure of some disk", they don't mean the mean time to failure of each individual disk but the mean time it takes until any one of the $100$ disks fails. If the failures form a Poisson process, then having $100$ disks instead of $1$ disk will increase ...


1

For $y<0$, obviously $f_Y(y)=0$. For $+\infty>y>4$, $$f_Y(y)=f_X(\sqrt{y})\bigg|\frac{dx}{dy}\bigg|$$ $$=\lambda e^{-\lambda(\sqrt{y}+2)}\frac{1}{2\sqrt{y}}$$ $$=\frac{\lambda e^{-2\lambda}}{2}\frac{e^{-\lambda\sqrt{y}}}{\sqrt{y}}$$ For $0 \le y \le 4$, ...


2

The answer depends on your choice of parametrization. The way you have characterized the survival function $$S_X(x) = \Pr[X > x] = \exp(-\alpha x^{\beta - 1}), \quad x \ge 0,$$ is a little different than other parametrizations. Without seeing the source document to determine how the authors themselves chose to parametrize the distribution, it is ...


1

If $X$ has probability density $f$, then the hazard rate function of $X$ is $h = \frac f{1-F}$ where $F$ is the distribution function of $X$. So we may compute for $t>0$ $$ h(t) = \frac{\alpha (\beta -1) x^{\beta -2} e^{-\alpha x^{\beta -1}}}{e^{-\alpha x^{\beta -1}}} = \alpha(\beta-1)x^{\beta-2}. $$


0

More generally, if $X$ and $Y$ are independent, with cdfs $F_X$ and $F_Y$, then $$ \eqalign{\Bbb P[X=Y] &=\sum_{x\in\Bbb R}\Bbb P[X=x]\cdot\Bbb P[Y=x]\cr &=\sum_{x\in\Bbb R}[F_X(x)-F_X(x-)]\cdot[F_Y(x)-F_Y(x-)],\cr } $$ so $\Bbb P[X=Y]=0$ provided one or the other has a continuous cdf. (Or simply if their cdfs have no common discontinuities.) For a ...


1

Suppose there are $n$ flowers. The number of flowers still blossoming at time $t$ is a random variable $X(t)$, equal to $$X(t)=\sum_{f=1}^n I(\text{flower $f$ is alive at time $t$})\tag1$$ where the indicator function $I(A)$ equals one if event $A$ is true, zero otherwise. The expectation of $X(t)$ is then $$E(X(t))=n P(\text{flower $1$ is alive at time ...


-1

Perhaps you should ask the following question: Fix $Z$ in your event space, what is $P(Y=Z)$? This might help you.


0

I have never encountered this concept before. Is this equation valid for $y>0$? $$\mathbb{P}(|X|>y) = \mathbb{P}(-|X|<y<|X|)$$ Yes, for all strictly positive $y$, then $-\lvert X\rvert < y <\lvert X\rvert$ is an equivalent event to $y<\lvert X\rvert $. However, whether this is helpful is another matter.   Does it simplify ...


1

First, randi([1 2]) will give you a uniform distribution from 1 to 2. And you said that you will count the difference between number of heads and number of tails. In general, if the coin-flip experiment is fair, the probability of getting a head should be the same as the probability of getting a tail = 0.5. That means the difference should be zero. But in ...


0

According to your linear transformation model $\Delta y = a \Delta x$ where $\Delta x = 1$. Therefore, $P(Y=k)$ is written as $$ P(Y=k) = P(Y \leq k) - P(Y \leq k - a), $$ $$ \hspace{0.75in} = F_x(\dfrac{k-b}{a}) - F_x(\dfrac{k-b-a}{a}), $$ $$ \hspace{0.75in} = F_x(\dfrac{k-b}{a}) - F_x(\dfrac{k-b}{a}-1), $$ Thus, we get $$ ...


2

We are told that $X = X_1+\dotsb+X_N$, where $N\sim\text{Pois}(\lambda)$. a) Essentially, you are asked to compute or give $P(X = k|N = n)$. If I tell you that $N = n$, then the sum of $n$ independent Bernoulli trials follows a binomial distribution with $(n,p)$. Hence, $X|N \sim \text{Bin}(n,p)$. b) I believe that you are essentially being asked to ...


2

Here is a guideline. Some details are left as smaller exercises, but I really encourage you to fill them up. I'll first recall how to prove that a sequence of random variables converges almost surely, before adapting the argument to the limsups. Convergence of sequences of random variables Let $(a_n)_{n \in \mathbb{N}}$ be a sequence of real numbers, and ...


0

Observe that if $X_1,\dots,X_n$ are independent, identically distributed (i.i.d.) random variables, all Bernoulli distributed with success probability $p$, then $$X = \sum_{k=1}^n X_k \sim \mathcal{B}(n,p) $$ that is $X$ has a binomial distribution. Your model is $Z|N\sim \mathcal{B}(N,p)$ is a binomial variable and $N$ is a Poisson variable $N\sim ...


0

Here is a possible reduction(?) of complexity, that someone with more experience may be able to turn into a solution. Consider the integral $$ I = \int_\gamma \; g^{-(a+1)}\;\exp\left\{-\left(\frac{b}{g} + \frac{1}{2} \sum_{i=1}^{n} \frac{t_i^2}{g+\lambda_i^{\psi}} \right) \right\} \log g \;\mathrm{d}g $$ Choose the branch cut of $\log g$ to run along the ...


1

1) Knowing that in the first four hours $8$ persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours. I define the random variable $Y_i$ =number of clients in the i-th hour of the opening hours $Y_i$=number of clients in the i-th hour of the opening hours for $i=1,2,3,4$ , since ...



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