New answers tagged

0

Output of $g^{-1}(A)$ is a set: $g^{-1}(A) = \{x \in \mathcal{X}: \ g(x) \in A\}$. this generalization into set mapping was built to be not contradicting with "ordinary" function $g(x)\leftrightarrow g(\{x\})$. For example, we can say, that $$g(\{x_1,x_2\})= \{g(x_1),g(x_2)\} = \{y_1,y_2\}$$ and $$g^{-1}(\{y_1,y_2\}) = \{x_1,x_2,x_3\}, $$ as there may ...


1

Both "Cov" and "Var" are used to represent the covariance matrix of the vector $\bf X$. See for example this Wikipedia remark: Nomenclatures differ. Some statisticians, following the probabilist William Feller, call the matrix $\Sigma$ the variance of the random vector X, because it is the natural generalization to higher dimensions of the 1-dimensional ...


0

Calculating probability of event 1 Given $r$ special objects labeled $o_1,o_2,\dots,o_r$ and $(n-r)$ additional ordinary objects, we ask what is the probability of when considering the special objects specifically that $o_1$ occurs before $o_2$ which occurs before any of the other special objects (with possibly some of the ordinary objects scattered ...


0

I don't think independence matters here...you should be able to do this with counting. Your sample space is the set of $n!$ permutations of your objects. Fix a value of $t_2 \le l$. The number of permutations which meet both of your conditions in this case is: $(t_2-1) \times \frac{(n-r)!}{(n-r-t_2+2)!} \times (n-t_2)!$ The $(t_2-1)$ is the number of ...


0

You have that $X_k\sim\mathcal N(1.8\textrm{cm},0.7^2\textrm{cm}^2)$ independently for all $\{X_k\}_{k\in\Bbb N^+}$ being the widths of books to be placed on a shelf of length $4600$cm. Now, let $Y_N=\sum_{k=1}^N X_k$ , and so $Y_N \sim \mathcal N(1.8\textrm{cm}\,N, 0.7^2\textrm{cm}^2\,N)$ $Y_N$ is thus the combined width of $N$ books. You want to find ...


0

In general, the Linearity of Expectation says that for any constants $a,b$ and random variable $X$, then: $$\mathsf E(aX+b) ~=~ a~\mathsf E(X)+b$$ If you can find $\mathsf E(X)$, then you can find $\mathsf E(X-c)$ and $\mathsf E(\tfrac 1 c X)$


1

I have a heuristic, very intuitive idea, so this is not a proof, just some thinking direction. Short: we possibly can take the conditional expectation over all the subsets of subalgebra $\sigma(C) \subset \mathcal{C}$, where $\mathbb{E}[X \mathbb{1}_{C}]\geq 0, \ C \in \mathcal{C}$ and set $Y=0$ elsewhere. At first, it is obvious that if $X$ takes only ...


0

a. $E(X)=\int_{c}^{\infty}xe^{-(x-c)}dx=\int_{c}^{\infty}xe^{-x+c}dx=-\left(x+1\right)\mathrm{e}^{c-x}\vert_{c}^{\infty}=0-(-(x+1))=x+1$ b. $E(X-c)=E(X)-c=x+1-c$ c. $E(X/c)=E(X)/c=\frac{x+1}{c}$ Hint: For the integral part $\int_{c}^{\infty}xe^{c-x}dx$, you could use integral by parts, and let $f(x)=x$ and $g'(x)=e^{c-x}$. Then $f'(x)=1$ and ...


0

There is an interesting geometric proof for this proposition given in Section $5$ of Probability and Measure by Patrick Billingsley. I will copy the illuminating figure from this textbook: Note that the equation to be proved can also be written as $$E[X] = \int_0^\infty P[X \geq x] dx. \tag{1} $$ For your special case, $x_i$s in the figure are integers ...


1

Call the one particular black ball we're concerned with $B$. Then the orders in question are: $$ \begin{array}{cccccccccl} W & W & W & \cdots & W & W & W & B & & w!\text{ different orders} \\ W & W & W & \cdots & W & W & B & W & & w!\text{ different orders} \\ W & W & W & ...


0

I came up with this: We have $\mathbf{x}^H\mathbf{y}=\mathbf{u}^H\mathbf{C_{xx}}^{\frac{1}{2}}\mathbf{C_{yy}}^{\frac{1}{2}}\mathbf{v}=tr\{\mathbf{C_{xx}}^{\frac{1}{2}}\mathbf{C_{yy}}^{\frac{1}{2}}\mathbf{v}\mathbf{u}^H\}\leq tr\{\mathbf{C_{xx}}^{\frac{1}{2}}\mathbf{C_{yy}}^{\frac{1}{2}}\}tr\{\mathbf{v}\mathbf{u}^H\}$ ...


2

They will have either 1 or 0 girls. Let $y$ be the probability that a child is a girl, then $E(Y) = 1-(1-y)^5$. $E(X) = 5 (1-y)^4 + 4 (1-y)^3y + 3 (1-y)^2y + 2 (1-y)y + y$


0

We say two numbers have the same order of magnitude of a number if the big one divided by the little one is less than 10. For example, 23 and 82 have the same order as their increment number can be found inside the numbers affiliated number by timing the 1-3 f magnitude, but 23 and 820 do not. -John C. Baez This is the definition of "Order of ...


2

The domain of $f(X)$, or actually $f\circ X$ is $\Omega$. It must be looked at it as the function $\Omega\to\mathbb R$ prescribed by $$\omega\mapsto f(X(\omega))$$ Defined like that also $f(X)$ is a random variable if $f:\mathbb R\to\mathbb R$ is a Borel-measurable function.


0

In general for random variables $U,V$ and constants $a,b,c,d$ we have the rules: $\text{Cov}(aU+b,cV+d)=ac\text{Cov}(U,V)$ $\text{Cov}(U,U)=\text{Var}(U)$ This preassumes of course that the expectations that are involved exist. Based on it we also find: $\text{Var}(aU+b)=a^2\text{Var}(U)$ Applying that here we find: ...


1

The statement is correct with $X_\tau$, but you have to read it carefully. The LHS is a conditional expectation, hence a random variable, so for some $\omega \in \Omega$ you have that $\mathbb{E}(f(X_{\tau+1}, \ldots) \mid \mathcal{F}_\tau) = \mathbb{E}_{X_{\tau}(\omega)}(f(X_0, \ldots))$, i.e. you can restart your Markov chain at some random point and ...


0

Random by itself doesn't mean much. There are 'patterns' or distributions of random (uniform, normal, Poisson, etc). Each distribution is a way of assigning a 'likelihood' or probability to certain events occuring in an experiment.


1

You can think of a sample space $\Omega$ as the set of all possible outcomes from some experiment. A sigma algebra $\mathcal{F}$ on $\Omega$ consists of events, which are collections of outcomes. A probability measure $P$ defined on $\mathcal{F}$ tells you how likely events are. One can think of a sigma algebra as containing information that is available, ...


7

Let $X$ be a nonnegative random variable with distribution function $F_X$. Then, by Fubini's Theorem, we have $$E[X]=\int_{[0,\infty)}\,x\,\text{d}F_X(x)=\int_{[0,\infty)}\,\int_{[0,\infty)}\,\chi_{[0,x)}(t)\,\text{d}t\,\text{d}F_X(x)=\int_{[0,\infty)}\,\int_{[0,\infty)}\,\chi_{[0,x)}(t)\,\text{d}F_X(x)\,\text{d}t\,,$$ where $\chi_E$ is the characteristic ...


7

$aP(X = a) = P(X=a) + \dots + P(X = a)$ ($a$ times) Now assuming $X$ takes only integer values $\ge 1$, you have $$1 = P(X > 0) = P(X = 1) + \color{red}{ P(X = 2)} + \color{green}{P(X=3) }\dots$$ $$P(X > 1) = \color{red}{P(X = 2)} + \color{green}{P(X = 3) }+ \dots $$ $$P( X > 2) = \color{green}{P(X = 3)} + P(X = 4) + \dots $$ and so on. ...


3

This article in Wikipedia gives an answer to your question. A slightly more general statement says the following: When a random variable $X$ takes only values in nonnegative integers, we can use the following formula for computing its expectation (even when the expectation is infinite): $$ \operatorname{E}[X]=\sum\limits_{i=1}^\infty P(X\geq ...


0

I think we can solve it in this way: $F_{E}(e)=P(E<=e)=\int_{-\infty}^{+\infty}ds_{1} \int_{-\infty}^{E-s_{1}}ds_{2} \int_{-\infty}^{E-s_{1}-s_{2}}f_{S_{1},S_{2},S_{3}}(s_{1},s_{2},s_{3})ds_{3}$ but I'm not sure. I like to know the solution too.


2

The error is that you have have assumed that the distribution of $\frac{V_1}{V_1+V_2}$ is distributed as $F(5,14)$. This is not the case, as $V_1$ and $V_1+V_2$ are not independent (have a look at the Characterisation section in here) - on the other hand, $\frac{V_1/5}{V_2/9}$ would be distributed as $F(5,9)$. Going back to the problem at hand, we can ...


3

I believe you should be differentiating with respect to $y$, not $x$, both in the first displayed equation, and in the subsequent derivation. You should also be stating the assumption that $g$ is a monotonic function. If it isn't, you need to be summing over different monotonic pieces of $g$ in your first displayed equation. You haven't shown the ...


3

The true roots of this formula lie in the change of variables using absolute value of Jacobian in calculus. Recall that in 2D case, if you transform a small square area $dS=dxdy$ into small area $dU=dudv$, then $$\int f(x,y)dxdy = \int h(u,v)\cdot \big|J(u,v)\big|dudv, \ \ \ \big|J(u,v)\big|=\bigg|\frac{\partial(x,y)}{\partial(u,v)}\bigg|$$ where $h(u,v) = ...


0

Here is a simulation using R statistical ssoftware of a million performances of this experiment, where $P(Heads) = .3$ for the biased coin. Results should be accurate to a couple of decimal places. You can use them as a 'reality check' for your work. m = 10^6; x = y = numeric(m) for(i in 1:m) { x[i] = rbinom(1, 2, .5) y[i] = rbinom(1, x[i], .3) } ...


0

First calculate out Cov(X,Y) using $Cov(X,Y)=\sum(X-\mu_X)(Y-\mu_Y)f(X,Y)$ where f(X,Y) is the correspondig pdf. Then use the formula of correlation coefficient: $cor(X,Y)=\frac{Cov(X,Y)}{\mu_X\mu_Y}$You can take a look at this example: https://onlinecourses.science.psu.edu/stat414/book/export/html/94 However, I think you made a few mistake in part a) and ...


1

$\mathsf P(X=x)$ is the probability that you select consecutively $x-1$ cards of one colour and then one card of the other colour; for $x\in\{2,3,4\}$ ~ out of all the ways to select any $x-1$ cards and then one more card.   (Which is not the same as just selecting any $x$ cards; the "last" position is held to be special.) $$\begin{align} \mathsf P(X=x) ...


1

Not quite what you're asking for, but a standard example of continuous random variables with the same moments yet distinct distributions is $f_1(x) = e^{- (\log{x})^2/ 2} \, / \,(x\sqrt{2\pi})$ $f_2(x) = f_1(x)(1 + \sin(2 \pi \log{x})/2)$ where $ x > 0$.


1

A fixed-effects model may be useful here. Let $t_{pcrn}$ be completion time for a given person $p$ driving in car $c$ on track $r$ for run $n$ (assuming there can be multiples of each triplet $pcr$. We could model $t_{pcrn}$ using a three-factor, additive, fixed-effects model: $$t_{pcrn} = \mu + \phi_p+\alpha_c + \beta_r + \epsilon_{pcrn}$$ Where: $\mu$ ...


1

Your value for $Pr(X=4)$ is incorrect. To make calculations easier, we may temporarily assume that each card has a number on it which makes it distinguishable from the others. I.e., we have a "red1", a "red2", etc... What it appears that you did was to count the number of ways in which you can pull four cards where order doesn't matter so that you get all ...


1

Let really count it by cases. We can only have $5$ cases. Note that they have different probability. $Pr[RRY]=3/5 \times2/4\times 2/3 =1/5$ $Pr[YYR]=2/5 \times 1/4\times 3/3 =1/10$ $Pr[RY]=3/5 \times 2/4=3/10$ $Pr[YR]=2/5 \times 3/4=3/10$ $Pr[RRRY]=3/5 \times 2/4 \times 1/3 \times 2/2 =1/10$ So we can see (also from the comment) the problem is from ...


2

If $X$ is independent from $Y$ and $Z$, is then $X$ independent from $YZ$? The answer is no. A counter example: take $Y,Z$ to be independent Rademacher random variables (i.e., they are uniform on $\{-1,1\}$), and set $X=YZ$. You can check that $X$ is itself a Rademacher random variable, and that $X,Y,Z$ are pairwise independent. But clearly, $X$ is ...


1

This is an inclusion-exclusion formula that sums over the number of gaps that the sequences might leave. You can perform the same type of calculation for your case. A complication arises because unlike in the continuous case, there's a finite probability that reads start at the same point, and this makes it more difficult to count the number of ways of ...


2

I doubt you'll get a solution in closed form for this, but in practice you can do a lot better than an exhaustive search over all possible orders. Build the orders one by one (e.g., start with just $AB$, then add $C$ in the three possible places, etc.) and abandon a branch once the probability for the part you've built is already lower than the probability ...


2

They want you to calculate the three unknowns. You could try to find three equations. They tell you that $E[Y] = 0$. Hence $$E[Y] = \sum_{k=-1}^1 k\cdot p_k= -1\cdot p_{-1}+0\cdot p_0+1\cdot p_1 = 0.\tag 1$$ They tell you that $\text{Var}(Y) = 1/2$, and hence $$\text{Var}(Y) = E[X^2]-\{E[Y]\}^2 = (-1)^2p_{-1}+0^2p_0+1^2p_1-(0)^2 = \frac{1}{2}.\tag 2$$ ...


0

Use the fact that $p(-1)+p(0)+p(1)=1 $ along with $E(X)=-p(-1)+p(1)=0$ and also $V(X)=E(X^2)-(E(X))^2 = p(1)+p(1) = \frac{1}{2}$


1

If $E[Y]=0$ what does that reveal about the probability that $Y=1$ and the probability that $Y=-1$. These two values should be equal, although we do not quite yet know what the probability necessarily is. So, let the probability $Y=1$ be $p$. $E[Y]=1\cdot p +(-1)\cdot p +0\cdot (1-2p)=0$ What about the variance? Now, $Var(Y)=1\cdot p+1\cdot ...


1

If $X$ is a uniformly distributed random variable over $[0,200]$, $$ \mathbb{P}[X^2+(200-X)^2 \geq 160^2]=\mathbb{P}[|X-100|\geq 20\sqrt{7}] $$ hence the wanted probability is: $$ 1-\frac{40\sqrt{7}}{200} = \color{red}{1-\frac{1}{5}\sqrt{7}} \approx 47\%. $$


1

First, we note that $P(A \ge t) = e^{-t/100}$ and $P(B \ge t) = e^{-t/200}$. Next, denote $X$ as the chosen lightbulb. We have \begin{align*} P(X = A|X \ge 200) &= \frac{P(X \ge 200|X = A) P(X = A)}{P(X \ge 200|X = A)P(X = A)+P(X \ge 200|X = B)P(X = B)} \\ &=\frac{e^{-200/100}\cdot \frac{3}{4}}{e^{-200/100}\cdot\frac{3}{4}+e^{-200/200}\cdot ...


1

You're mixing up units. If you define the interval as $[0, 2]$, then you need to set the inequality $\geq 1.6$ (as pointed out by @N74's comment). Alternatively, if you keep the units in cm, then the interval is $[0, 200]$. You then have the inequality $X^2 + (200 - X)^2 \geq 160^2$ (or $X^2 + (2 - X)^2 \geq 1.6^2$). Simplify the inequality and solve for $X$ ...


1

Call pulling a white ball a "success." We are repeating an experiment independently $3$ times, where each time the probability of success is $2/5$. The random variable $X$ gives the number of successes in $3$ independent trials. Thus $X$ has binomial distribution, parameters $n=3$, $p=2/5$. For $k=0,1,2,3$, we have ...


0

Sorry, I'm not good with operating probabilities through densities and integrals, so this solution might be incorrect, but you'll get the idea: I will reformulate the task as "John starts to eat between 11 and 12.45", etc. Then we have that random variable $Y$, uniformly distributed on $[15,120]$ (in minutes) should get its value between $X$ and $X+45$, ...


2

Statement 1 follows from the definition of $\mathbb{E}(X|Y)$: A version of $\mathbb{E}(X|Y)$ is a variable $Z$ such that $Z$ is $\sigma(Y)$ measurable and $\mathbb{E}(X1_A)=\mathbb{E}(Z1_A)$ for all $A\in\sigma(Y)$. It is trivial that $\Omega\in\sigma (Y)$ (where $\Omega$ is the sample space), hence ...


1

I'm not sure, I do not like the proof of (1) as I don't remember that existence of densities is necessary. Lets begin with $\sigma$-algebras, just using the fact that for random variable $Y$ by definition $\mathbb{E}[X|Y]= \mathbb{E}\mathbb[X|\sigma(Y)]$. I will take the following definition of conditional expectation with respect to sigma-algebra: ...


1

Since $X_n \to X$ a.s, there's a set $A$, $P(A)=1$ such that for every $\omega\in A$, $X_n(\omega) \to X(\omega)$. Therefore (calculus) on $A \cap \{\omega:X(\omega)\ne 0\}$ we have $\frac{1}{X_n(\omega)}\to \frac{1}{X(\omega)}$. Summary: Special case. If $X\ne 0$ a.s., then indeed $\frac{1}{X_n}\to \frac{1}{X}$ a.s. General. $\frac{1}{X_n} \to ...


2

Assume $X\sim U(0,2)$, that is the cut point is uniformly distributed between $0$ and $2$. Then if $X=x$ the area $A=a=x(2-x)$ and this is greater than a half between the roots of $x(2-x)=1/2$ which are $1\pm\sqrt{1/2}$. The pdf of $X$ is $f_X(x)=1/2$ for $x\in (0,2)$ and zero otherwise. So the probability that$A\ge 1/2$ is: $$ P(\mbox{A} \ge ...


3

If you solve $X(2-X)<0.5$ for $X$ you get $X>1+\frac{1}{\sqrt{2}}$ and $X<1-\frac{1}{\sqrt{2}}$ The cdf of $X$ is $F(x)=\begin{cases}0, \ \ x<0 \\ \frac{x}{2}, \ \ \ 0\leq x\leq 2 \\ 1, \ \ x>0\end{cases}$ It is $F(X>x)=1-F(X<x)$ Therefore you have to calculate $F(1-\frac{1}{\sqrt{2}})+(1-F(1+\frac{1}{\sqrt{2}}))$ Remark The ...


1

Well, correlated or not, the distribution of X1+X2 is going to be normal, so I don't see any big difference. Only the variance of this is going to change according to the correlation.


2

We have $$\mathbb P\left(\bigcap_{\substack{a\lt b\\a,b\in\mathbb Q^+}}\bigcup_{n\geqslant 0}\left\{X_n\in (a,b)\right\}\right)=1$$ because for each $a,b\in\mathbb Q^+$ such that $a\lt b$, $$\mathbb P\left(\bigcap_{n\geqslant 0}\left\{X_n\notin (a,b)\right\}\right)\leqslant \mathbb P\left(\bigcap_{n= 0}^N\left\{X_n\notin (a,b)\right\}\right)=\left(\mathbb ...



Top 50 recent answers are included