# Tag Info

4

Consider $\Omega = [-1, 1]$, $X(t) = t$, and set the codomain of $Y$ to $\{0, 1\}$, with $$Y(t) = \begin{cases} 0 & t \le 0 \\ 1 & \text{else} \end{cases}.$$ Let $x = 0$ and $y = 0$ in your last paragraph. The first set is $\{0\}$ and the second is $[-1, 0]$, and their intersection is $\{0\}$. So the answer is "no" to that main question. It ...

4

If I am not mistaken, $\left\{Y=X \right\}$ and $\left\{Y \neq X\right\}$ partition the space. So $\left\{X=m \right\} \subset \left\{ Y=m \right\} \cup \left\{Y\neq X\right\}$ and the symmetric statement (change the roles of $X$ and $Y$) give your inequality.

3

"Theory of Stochastic Processes", Gusak et al., Springer, 2010: Problem 1.10: Prove that it is impossible to construct on the probability space $\Omega = [0, 1]$, $\mathcal{F}=\mathcal{B}([0,1])$, $\mathsf{P}=\lambda$ a family of independent identically distributed random variables $\{\xi_t, t\in[0,1]\}$ with a nondegenerate distribution. ($\lambda$ ...

2

You can solve it by inspection.   Just fill in the table: $$\begin{array}{l:l}x & \dfrac{2^x~\mathsf e^{-2}}{x!} & \displaystyle\sum\limits_{k=1}^x\dfrac{2^k~\mathsf e^{-2}}{k!} \\ \hdashline 0 \\ 1 \\ 2 \\ 3 \\ 4\\ 5 \\ \hline 6 \\ \vdots \end{array}$$ Stop when the third column exceeds $0.99$.   Do you have access to a spreadsheet ...

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 $0 \le y \le 4$, ...

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.

1

Hint:$$\Phi^{-1}(U)\leq x\iff U\leq \Phi(x)$$ Here $\Phi$ denotes the CDF corresponding with standard normal distribution.

1

No matter what values all except the last variable are assigned, there is exactly one value the last variable can be assigned to satisfy the equation. Since all choices are independent, the probability of satisfying the equation is the same as the last variable taking the correct value, which happens with probability $\frac{1}{2}$.

1

It is not that complicated. The conditional expectation, of random variable given an event, is defined in terms of conditional probability. \begin{align}\mathsf E(Z\mid Z\geq 0) ~=~& \int_\Bbb R z~f_Z(z\mid Z\geq 0)\operatorname d z\\[1ex]~=~& \int_0^1 z f_Z(z)\operatorname d z~\Big/~\int_0^1 f_Z(z)\operatorname d z \end{align} Also you have ...

1

Let $R$ be the distance between $O$, the origin, and $M$. The probability that $R$ is less than or equal to a value $r$ is $$P(R\le r) = \begin{cases} \frac{\pi r^2}{\pi\cdot 1^2} = r^2, & 0\le r\le 1\\ 1, &r>1\\ 0, &\text{otherwise} \end{cases}$$ The probability density function of $R$ is $$f_R(r) = \frac{d}{dr}P(R\le r) = \begin{cases} 2r, ... 1 I started writing this before peterwhy posted his answer, and since its approach is a little different I decided to post it anyway. We consider the problem on the unit disk \mathbb{D} of determining \mathbb{P}(\text{d(M,N) \leq 1}). To that end, we shall use the law of total probability. Rather than consider a single partition of our sample, we ... 1 Clearly, the stopping criterion is that the second can of tomatoes is opened. If there are three cans of peas, then it is entirely possible that all three cans could be opened before the second can of tomatoes is opened, since the final can that is selected could be the second tomato can. The most direct way to compute the joint distribution is to consider ... 1 2 cans of tomatoes, 3 of peas and 1 of beans Y is the number of cans opened until the 2 cans of tomatoes are open. We measure the probability of selecting 1 from 2 tomato cans and y-2 from the 4 others in some arrangement out of the ways to select y-1 from 6 cans, times the conditioned probability of selecting the 1 tomato can from the ... 1 1) The conditional distribution of the 8 clients along the interval [0,4] is Uniform, so the conditional distribution of the number of clients in a sub interval (2,4] is Binomial with p=\frac{2}{4} and n=8. Denote it by X, the event of interest is \{X=4\}, so P(X=4) = \binom{8}{4}\frac{1}{2^8}. The answers to 2 and 3 look OK to me. 1 1) Knowing that in the first four hours 8 persons got into the bank, calculate the probability of exactly half of them having entered past the first two hours. I define the random variable Y_i =number of clients in the i-th hour of the opening hours Y_i=number of clients in the i-th hour of the opening hours for i=1,2,3,4 , since ... 1 I often don't know what to do, besides using somehow Borel-Cantelli Well, more often than not, you don't need to use anything else:) As for the question:$$ \sum_{n=0}^{\infty}P(|X_n|^p\geq n)=\sum_{n=0}^{\infty}P(|X_1|^p\geq n)\geq \int_0^{\infty}P(|X_1|^p\geq t)dt=E|X_1|^p=\infty. $$1 If the students have mean \mu and variance \sigma^2, then the distribution for the average of n students is approximately normal with mean \mu and variance \frac{\sigma^2}{n}. In this case, you want to know the probability that the sample mean \bar{X} is within \pm r of the true mean \mu. So,$$P(\mu-r \leq \bar{X} \leq \mu-r) = P(-r \leq ...

1

You've shown that there is a $c$ such that $P(T=\infty)>0$. So:$$E[A_{T\wedge n}]=E[A_{T\wedge n}|T<\infty]P(T<\infty)+A_n P(T=\infty).$$ Both terms terms on the r.h.s are non-negative and you've also shown that $E[A_{T\wedge n}]$ is finite and you can take $n\rightarrow\infty$, so at the very least $A_\infty<\infty$ since $P(T=\infty)>0$

1

We can evaluate $\operatorname E(XY)$ using the law of total expectation. We have that $$\operatorname E(XY)=\operatorname E(\operatorname E(XY\mid X))=\operatorname E(X\operatorname E(Y\mid X))=\operatorname E(X^2+X)=\operatorname EX^2+\operatorname EX$$ and $$\operatorname{Cov}(X,Y)=\operatorname E(XY)-\operatorname E(X)\operatorname E(Y)=\operatorname ... 1 (X,Z) are independent if forany f,g$$\mathbb{E}(f(X)g(Z))=\mathbb{E}(f(X))\mathbb{E}(g(Z))$$In your case :$$\mathbb{E}(f(X)g(Z))=\mathbb{E}\left(f(X)g\left(\frac{Y-pX}{\sqrt{1-p^2}}\right)\right)=\int_{\mathbb{R}^2}f(x)g\left(\frac{y-px}{\sqrt{1-p^2}}\right)p(x,y)~dx~dy$$so this is a change of variable problem, you set : ... 1 Hint: You can write$$ \mathbb{E}(X\mid X\geq t)=\frac{\int_t^\infty xf_X(x)\,dx}{\int_t^\infty f_X(x) \, dx}.  Using this to get a differential equation for $f_X$.

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