Tag Info

Hot answers tagged

5

First, $$P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c)=E[P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c|X_2,X_3)].$$ As far as the conditional probability: $$P(X_1+X_2 \leq a, X_2+X_3 \leq b, X_3 +X_4 > c|X_2=x_2,X_3=x_3)=$$ $$=P(X_1 \leq a-x_2, x_2+x_3 \leq b, X_4 > c-x_3|X_2=x_2,X_3=x_3)=$$ $$ =\begin{cases} 0, & \text{ if } & ...


2

The probability that two independent continuous random variables are exactly equal is 0. And, of course, normal random variables are continuous. Continuous random variables put 0 probability on any one point. A more interesting problem might be the probability that the two variables are within one unit of each other. For discrete random variables the ...


2

$\newcommand{\E}{\operatorname{E}}\newcommand{\var}{\operatorname{var}}$ It follows from independence that $\E(hk)=\E(h)\E(k)$. If $\E(h)=\E(k) = 0$ then $\E(h^2)= \var(h)$ and $\E(k^2)=\var(k)$ and we have \begin{align} \E\left((\sqrt{t}\, h + k)^2\right) = \E(th^2 + 2\sqrt t\,hk + k^2) & = t\E(h^2) + 2\sqrt t\E(h)\E(k) + \E(k^2) \\[10pt] & = ...


2

Let $W$ (for example) be a random variable which takes on values $0$ and $1$, each with probability $\frac{1}{2}$. let $Z$ be standard normal, and suppose $W$ and $Z$ are independent. Let $X=W+Z$ and let $Y=W$. Then $X-Y$ is normally distributed, but neither $X$ nor $Y$ is normal. Remark: Note that if $X$ and $Y$ are normal and not independent, then $X-Y$ ...


2

On 1) Let $W:=X+Y$. Then: $$F_{W}\left(w\right)=P\left(X+Y\leq w\mid X=0\right)P\left(X=0\right)+P\left(X+Y\leq w\mid X=1\right)P\left(X=1\right)=$$$$\frac{1}{2}F_{Y}\left(w\right)+\frac{1}{2}F_{Y}\left(w-1\right)$$ Here $F_{Y}$ is well known to you and knowing CDF $F_{W}$ you can find PDF $f_{W}$. On 2) $X=0\Rightarrow XY=0$ so that $P\left\{ ...


1

Using the the multinomial theorem $$\frac{400!}{k!\ l!\ (400-k-l)!}(0.5)^k(0.2)^l(0.3)^{400-k-l}$$ is the probability of any combination. You want to sum all $(k,l)$ pairs such that $k+2l\leq 390$ This will be difficult to actually evaluate, you might want to use approximations.


1

Notice that the pdf of $X$ is $$f_X(x)=\begin{cases} \frac{1}{18},&\text{ if } -9\le x \le 9\\ \ 0,& \text{ otherwise. }\end{cases}$$ We can treat $F_Y(y)=P(Y<y)$ as the sum of two probabilities: $$F_Y(y)=P(Y<y\cap |X|>3)+P(Y<y\cap |X|\le 3).$$ First, $$P(Y<y\cap |X|>3)=P(0<y \cap |X|>3)=\begin{cases} 0,&\text{ if } ...


1

One can use product probability space to model a Poisson process. Let $$\Omega = \prod_{k = 1}^\infty \mathbb{R}^+, \mathcal{F} = \sigma\left(\prod_{k=1}^\infty\mathcal{B}(\mathbb{R}^+)\right)\text{ and }\mu = \prod_{k=1}^\infty \mathrm{Poisson}(\mathbb{R}^+),$$ where $\mathcal{B}(\mathbb{R}^+)$ is the Borel $\sigma$-algebra on $\mathbb{R}^+$ and ...


1

Let $W=\min(|X-1|, 5-X)$. We find the cdf of $W$. The cdf of $Y$ can then be obtained easily by pushing everything forward by $1$. Apologies for the length. I wanted to put in lots of detail, and do not have time to tighten things up. Draw carefully the curves $y=|x-1|$ and $y=5-x$. Everything depends on that. Note that $y=|x-1$ is $y=1-x$ to the left of ...


1

Yes.   That the probabilities of those four data points sum to one indicates that they are most certainly all of the possible outcomes.   Then: $$\mathsf P\{X\leq 3\} = \mathsf P_{X,Y}(1,1)+\mathsf P_{X,Y}(2,2)+\mathsf P_{X,Y}(3,3)\qquad\color{red}{\checkmark}$$


1

You can first simplify the problem: Let 1 denote an odd number and 0 and even number. Case 1: P(X=0) $11111$ $00000$ The even number can be arranged in 5! ways. The odd numbers can be arranged in 5! ways, too. Thus you have the factor $(5!)^2$ in all cases. In each coloumn 0 can be in the first row or in the second row. This gives an additional factor of ...


1

Let us grind it out. There are five odd and five even. So we might as well assume that we use five $0$'s and five $1$'s. Let $X$ be the number of columns with two $0$'s. We want the probability distribution of $X$. The random variable $X$ can only take on the values $0$, $1$, and $2$. So we really have only two probabilities to compute, since if we know ...


1

Both averages are calculated correctly. They differ because there are two sampling models being employed -- in one approach we sample a class at random and ask what is the average; in the other approach we sample a student at random and ask what is the average class size the student sees. As for which number you would choose to represent the 'average' class ...


1

Take some $Z$ following a normal distribution, and $X$ any other. Then set $Y=X-Z$. $X-Y$ follows a normal distribution, while $X$ doesn't.


1

Let $N \in\{1,2, \cdots ,n\}$ denote the number of attempts. The conditional probability we are looking for is $$P(N=k|N\le n)=\frac{P(N=k\cap N\le n)}{P(N\le n)}.$$ First, $$P(N=k\cap N\le n)=\begin{cases}0,& \text{ if }& k>N\\ P(N=k)=(1-p)^{k-1}p,& \text{ otherwise. } \end{cases}$$ Second, $$P(N\le n)=\sum_{i=1}^n(1-p)^{i-1}p.$$ So, for ...


1

Building on Michael Hardy's answer. I use the notation $\bar h = E(h)$ and $\bar k = E(k)$ for readability. Since $h,k$ are independent, $E(hk) = \bar h \bar k$. By definition, $E(h^2) = \sigma_h + \bar h^2$ and $E(k^2) = \sigma_k + \bar k^2$. We have, $$\begin{align*}E((\sqrt{t}h+k)^2)& =E(th^2+2\sqrt{t}hk + k^2)\\ &= ...


1

For reference, I am repeating the PDF (or PMF) table of $X_2.$ Then $E(X_2) = \sum_{i=0}^6 v_i p_i,$ the terms and total of which are in the third row. (My notation may different from that of your book, so make sure you connect the table with whatever formula your book has.) value 0 0 1 1 0 0 1 1 Total prob. .04 .08 .08 ...


1

It looks like you're asking why the variance and expected value of a boolean variable isn't itself boolean. The problem goes away if you consider variance to be a measure of the variable rather than the variable itself, and consider 'expected value' to be a form of scaled probability rather than the domain value to 'expect'. For example, given instances ...


1

It's no different than an integer variable with values $\{0,1\}$. Naturally, both EV and variance will be in the real interval $[0,1]$. What you have to do is simply find the distribution of $X_2$ from the table - this will be parameterized by $p\in[0,1]$ with $P(X_2=1) = p$ and $P(X_2=0)=1-p$. The expected value will then be $\mathbb E(X_2) = p$ and the ...


1

Since the the map $1-|x|$ is not invertable on $[-1,1]$, you cannot apply the general formula, you have to do some calculations instead: a) Integrate over $[-1,y]$ to get the CDF: $$\int_{-1}^y \frac{1-x}{2}\mathrm{d}x=C+\frac{2y-y^2}{4}$$ As it is a CDF, $C=\frac{3}{4}$, so $$F_X(x)=\frac{3+2x-x^2}{4}$$ b) $|X|\leq 1$, so: ...


1

$\begin{align} P(XY<1) & = \int_0^3 P(X<1/y \mid Y=y)f_Y(y)\operatorname d y \\ & = \int_0^3 \int_0^{\min\{2, 1/y\}} f_{X,Y}(x,y)\operatorname dx \operatorname d y & \tag{$\star$} \\ & = \frac 1 6\left( \int_0^{1/2} \int_0^{2} 1\operatorname dx \operatorname d y + \int_{1/2}^3 \int_0^{1/y} 1\operatorname dx \operatorname d y \right) ...


1

See the figure: You must calculate the blue area and divide it by the full area ($2 \times 3 = 6$). You should find: $P[x\ y < 1] = \frac{1}{6} (1+\log (6))$.



Only top voted, non community-wiki answers of a minimum length are eligible