Tag Info

Hot answers tagged

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 ...



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