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6

It can be shown that nonnegative random variables $X$ and $Y$ have the same distribution so long as $\mathbb{E}[X^\alpha]=\mathbb{E}[Y^\alpha]$ is finite for all $\alpha\in(a,b]$, any $0\le a <b$. Setting $U=\log X$ and $V=\log Y$, define the functions $$f(\alpha)=\mathbb{E}[1_{\{X > 0\}}e^{\alpha U}],\ g(\alpha)=\mathbb{E}[1_{\{Y > 0\}}e^{\alpha ... 5 Step I: Pick 6 balls and weigh them 3 on each side. Step II: 1. If the balls balance each other, then weigh the remaining two balls and the heavier one will tilt the balance on its side. 2. If the balls do not balance each other, then the heavier ball should be one of the three balls on the side the balance tilts. From these three balls, ... 3 You have three distinct boxes, and want to count ways to put 20 indistinct balls into them. The total count of solutions is, as you have calculated:$$\frac{22!}{2!20!} = 231$$Now, to generate forbidden solutions you can choose 2 boxes, then put the same number of balls n into each of them, and the remainder into the other. Since the number n can vary ... 3 Here's a hands-on intuitive approach to the problem. Suppose we have an alphabet \mathcal{A}=\{a_1,\ldots,a_k\}. Given nonnegative integers n_1,\ldots,n_k and m_1,\ldots, m_k, let$$\left[\begin{array}{c}n_1,\ldots,n_k\\m_1,\ldots,m_k\end{array}\right]$$denote the number of words that can be formed from n_i copies of a_i, such that the word ... 3 Let$$F(x):=x^{13}\cdot(\frac{(-x)^0}{3!}+\frac{(-x)^1}{2!}+\frac{(-x)^2}{1!}+\frac{(-x)^3}{0!})^{13}$$Then$$4!^{13}\cdot\sum_{k=13}^{52} \frac{1}{k!}[x^k]F(x)=\frac{50972203946555791528902451677555189167087762981}{92024242230271040357108320801872044844750000000000} =0.000553899741\cdots$$is the required probability. Ignoring suits, the number of ... 3 Hint: Condition on the result of the coin-flip. 3 If P(A)>0 then P(B|A)=\frac{P(A \cap B)}{P(A)}, as you saw in elementary probability. So if X=\chi_C then$$\int_\Omega X(\omega) dP(\omega|A) = \frac{P(C \cap A)}{P(A)} = \frac{1}{P(A)} \int_A X dP.$$Now extend to simple functions and finally random variables (as usual). 2 Hints: The \sigma-algebra \sigma(Y,Z) is generated by sets of the form$$\{Y \in A\} \cap \{Z \in B\}$$for Borel sets A,B \in \mathcal{B}(\mathbb{R}). By step 1 and the definition of conditional expectation, it suffices to show that$$\int_{\{Y \in A\} \cap \{Z \in B\}} \mathbb{E}(X \mid Y) \, d\mathbb{P} = \int_{\{Y \in A\} \cap \{Z \in B\}} X \, ...

2

For any $n \ge 0$, let $$f_n = {\bf Pr}\big[ S_0 =n \land \exists t > 0, S_t = 0 \big]$$ be the probability for a random walk start at location $n$ returns to origin in some finite time $t$. Let $q = 1-p$ and $\displaystyle\;\mu = \frac{q}{p}\;$. It is easy to see $f_n$ satisfies a recurrence relation $$f_n = \begin{cases} 1, & n = 0,\\ p f_{n+2} ... 2 I claim that$$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;E\big[XE[Y|\mathcal{G}]\big]=E\big[E[X|\mathcal{G}]\cdot E[Y|\mathcal{G}] \big]\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(*)If we can prove this then we will be finished, because the right-hand side of equation (*) is symmetric in X and Y, and so the ... 2 Three-dimensional case (n=3) The Wikipedia article on area and volume element states that the area element of the unit sphere is \sin\phi\,\mathrm d\phi\,\mathrm d\theta where I use \phi to denote inclination and \theta to denote azimuth. If you consider your fixed vector as the zenith direction, then my  \phi will be the same as yours. To turn ... 2 I believe that \{\tau\leq 1\} is a shorthand for \begin{align*} \{\omega\in\Omega\,|\,\tau(\omega)\leq 1\}. \end{align*} To see that this is consistent with the statement that \{\tau\leq 1\}=\{1\}, note that \begin{align*} \tau(1)=&\,\inf\{t\geq 0\,|\,\max\{t-1,0\}>0\}=\inf\{(1,\infty)\}=1,\\ \tau(2)=&\,\inf\{t\geq ... 2 Z_i and W aren't needed: you just need to note \forall i, E(X_i^2)=\text{Var}(X_i)+E(X_i)^2=\alpha+\alpha^2 and use the linearity of expectation to get: E\left[\sum_{i=1}^n X_i^2\right]=\sum_{i=1}^nE(X_i^2)=n\alpha(1+\alpha). $$Note also that the independence assumption is superfluous. 2 For any 0 = t_0 < \ldots < t_n, we know that$$B_{t_n}-B_{t_{n-1}}, B_{t_{n-1}}-B_{t_{n-2}},\dots,B_{t_1}-\underbrace{B_{t_0}}_{0}$$are independent random variables. Choosing t_0 = 0, t_1 = 1, t_2 = 2, t_3 = 3, t_4 = \pi, t_5 = 4, we get that$$B_4-B_{\pi},B_{\pi}-B_3, B_3-B_2,B_2-B_1,B_1 \tag{1}$$are independent. This implies in ... 2 A finite-dimensional distribution is just the joint distribution of a random vector. You pick various time points t_1,t_2,\ldots,t_n, observe the stochastic process at those time points, and ask yourself "what's the distribution of this random vector?" To define a FDD it is sufficient to specify the joint distribution on rectangles A_1\times A_2\times ... 2 Some basic identities P(A\cup B)= P(A) + P(B) - P(A\cap B) P(A-B) = P(A\cap B') = P(A) - P(A\cap B) P(A')= 1-P(A) P(A\cup B)= 1-P\big((A\cup B)'\big) = 1- 0.2 = 0.8 Then, P(A\cap B) = P(A) + P(B) - P(A\cup B) = 0.2  Thus, P(A\cap B')= P(A) - P(A\cap B) = 0.5-0.2 =0.3 Second edit Yes, this cannot be true for the following reasons. One ... 2 U is uniformly distributed hence f_u \,du = 1 \, du. The function you are taking the expected value of is x_t (U) = \sin(2\pi Ut). Thus, you want \Bbb E_U[x_t] = \int_0^1 x_t f_u \,du = \int_0^1 \sin(2\pi Ut)\cdot 1\, du For further information, look up the box muller method for simulating normal random variables. It's very similar. 2 Let A be the event that \int_0^\infty e^{B_t} \,dt < +\infty. By the Kolomorgov 0-1 law, P(A) = 0 or 1. Now let B be the event that \int_0^\infty e^{-B_t}\,dt < +\infty. By symmetry, P(A) = P(B). Moreover, on A \cap B we have$$\int_0^\infty (e^{B_t} + e^{-B_t}) \,dt < +\infty$$which is absurd since e^x + e^{-x} \ge 1 for all ... 2 Let A the event: the letter comes from TATANAGAR, B the event: the letter comes from CALCUTTA. You have:$$ P(TA | A) = \frac 28 = \frac 14 \\ P(TA | B) = \frac 17 $$Assuming that P(A) = P(B) Up to here I agree with mookid. P(A)=P(B)=\frac 12 Law of total probability:$$P(TA)=P(TA|A)\cdot P(A)+P(TA|B)\cdot P(B)=\frac 14 \cdot ...

2

The coin tosses can be modelled as a sequence of independent $\text{Ber}\left(\frac{1}{2}\right)$ random variables $(X_n)_{n\in\mathbb{N}}$ Letting $A_n=\{X_n=1\wedge X_{n+1}=1\}$, you want to know $P(\lim\sup_n A_n)$ Consider $\lim\sup_nA_{2n}\subseteq\lim\sup_nA_n$. Notice that $(A_{2n})_{n\in\mathbb{N}}$ are independent. Since ...

2

\begin{align*} \frac13=&\,0.010101\ldots\\ \frac23=&\,0.101010\ldots\\ 1=&\,0.111111\ldots \end{align*} in binary. $1/3$ and $2/3$ are normal, $1$ is not. If you insist on using translation invariance, suppose—for the sake of contradiction—that sums and differences of normal numbers are always normal. This implies that if $x$ is not normal and ...

2

The claim holds only true if $X$ is assumed to be real-valued, i.e. $\mathbb{P}(|X|<\infty)=1$. If this assumption holds true, then the continuity of the measure $\mathbb{P}$ implies $$\mathbb{P}(|X| \geq r) \leq \delta$$ for $r>0$ sufficiently large where $\delta$ is chosen as in the assumption for $\epsilon := 1$. Hence, by assumption, ...

2

You are correct in deriving the marginal pdf of $X$ and the conditional pdf of $Y$ given $X=x$. Now, for the last part, you want to integrate the joint pdf of $X$ and $Y$ over all points $(x,y)$ such that $x+y\geq 1$. We have \begin{align*} P(X+Y \geq 1) &= P(Y \geq 1-X) \\ &= \int\limits_{\frac12}^{1} \int\limits_{1-x}^{x} 10x^2 y \;\mathrm{d}y ...

2

Finite additivity and countable sub-additivity is equivalent to countable additivity. The proof is below. Let $\mu$ be a measure. It is clear that if $\mu$ is countably additive then it is finitely additive and countably sub-additive. Assume that $\mu$ is countably sub-additive and finitely additive. Consider a collection $\{A_n \}_{n=1}^\infty$ of ...

2

Denote by $$p(t,x) := \frac{1}{\sqrt{2\pi t}} \exp \left(- \frac{x^2}{2t} \right), \qquad x \in \mathbb{R},$$ the density of the normal distribution with mean $0$ and variance $t$. As you already noted, this function solves the heat kernel equation, i.e. $$\frac{\partial}{\partial t} p(t,x) = \frac{1}{2} \frac{\partial^2}{\partial x^2} p(t,x).$$ For $f ... 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 ... 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 $$\operatorname{E}[n/T] = \int_{x=0}^\infty \frac{n}{x} ... 1 first you can get b simply be computing the time zero expectation:$$\mathbb{E}(e^{5B_t})= e^{0.5 \times 25 \times t}.$$So b = 12.5. With this value, your final equation$$\exp\{\frac{25(t-s)}{2}+5B_s-bt\} = \exp\{5B_s-bs\}$$holds and we are done. 1 It is not certain that 6 people will be dead and 4 alive. That is the most likely outcome, but, you could imagine a very unlucky month where each person died. The likelihood for each number of deaths is plotted below. 1 Your MLE for \theta does not seem correct. If \boldsymbol x = (x_1, \ldots, x_n) is an iid sample drawn from a X \sim \operatorname{Pareto}(1,\theta) distribution with density$$f_X(x) = \theta x^{-\theta-1} \mathbb{1}(x > 1)$$then the likelihood function is$$\mathcal L (\theta \mid \boldsymbol x) = \theta^n \left(\prod_{i=1}^n x_i ...

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