# Tag Info

3

Knowing that the 6th coin flip is still 50-50 and completely independent from the last 5 should make the entire intro of your question void. You should ask - is it worth borrowing \$100 in order to have a 50-50 chance of winning an additional \$100. Everything else is just just noise. To answer this just think about the simple goal of maximising the ...

1

I would say you should not. Considering that the loan shark surely charges an unreasonably high interest rate, your expected value for this sixth flip is negative. Don't do it. And if you lose and can't repay the shark, you may end up with hospital bills too from the broken legs. Just go back to working hard and try to make some more money.

4

This answer is under the assumption that these are fair dice. You have $P(x_2 - x_1 = k) = (6-|k|)/36$, for $-5 \le k \le 5$. (This is just a short way of writing what you'd get out of counting case by case.) Now $$P((x_2 - x_1) = (x_4 - x_3)) = \sum_{k=-5}^5 P((x_2 - x_1) = k, (x_4 - x_3) = k)$$ and since $x_1, x_2$ are independent of $x_3, x_4$, this ...

2

The difference between the terms "probability measure" and "probability distribution" is in some ways more of a difference between terms rather than a difference between the things that the terms refer to. It's more about the way the terms are used. A probability distribution or a probability measure is a function assigning probabilities to measurable ...

0

Let's prove it for a nonnegative $X$. A stepfunction $0\leq f\leq X1_A$ can be written as $c_1\times 1_{B_1}+\cdots+c_n\times 1_{B_n}$ with $c_i>0$. This leads to $B_i\subseteq A$ so that $0\leq P(B_i)\leq P(A)=0$. So $\int fdP=c_1\times P(B_1)+\cdots+c_n\times P(B_n)=0$. Then $\mathbb EX1_A=0$ as supremum of these values.

0

Proposition: Let $(\Omega,\mathcal{F},P)$ be a measure space, $f:\Omega\to [0,\infty)$ a measurable function and $A$ a measurable set with $P(A)=0$. Then, $$\int_AfdP=0.$$ Proof: First, suppose $s=\sum_{i=1}^n a_i\chi_{A_i}$ is a simple measurable function (i.e. the range of $s$ is a finite set $\{a_1,\ldots,a_n\}$, and ...

2

Since $P(X\in A\mid Y) = P(Z\in A\mid Y)$ a.s. we have $E[P(X\in A\mid Y)\cdot 1_{\{Y\in B\}}] = E[P(Z\in A\mid Y)\cdot 1_{\{Y\in B\}}]$. Using the tower property of the conditional expectation and $\sigma(Y)$-measurability of the indicators, you now get the desired result.

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For any measure space $(X,\mathcal M,\mu)$, $\mu(\varnothing)=0$ by definition. Since a probability measure is just a special case where $\mu(X)=1$, we still have $\mu(\varnothing)=0$.

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Hint: Note that $A\cup \emptyset = A$, thus: $P(A\cup \emptyset) = P(A)+P(\emptyset)+P(A\cap\emptyset) = P(A)$, now $P(A)>0 \implies ?$

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Hint: $$P(\emptyset)=P(\emptyset\cup\emptyset)=P(\emptyset)+P(\emptyset)$$ This because $\emptyset\cap\emptyset=\emptyset$, i.e. the sets are disjoint.

0

We can use a symmetrization argument: take $(X'_i)_i$ and $(X''_i)$ two independent i.i.d. sequences, where $X'_1$ and $X''_1$ have the same distribution as $X_1$. Using Lévy's inequality $$\mathbb P\left(\max_{1\leqslant j\leqslant n}|S_j|\gt x\right)\leqslant 2\mathbb P( |S_n|\gt x) ,$$ we get that $$\frac 1{a_n} \max_{1\leqslant j\leqslant ... 0 The expectation of X_i does not necessarily exist, but even if it does, the sequence of expectations may not converge to 0. For example consider a probability measure \mathbb P on the set of positive integers such that \mathbb P\{n\}= 2^{-n} (here \Omega=\mathbb N\setminus\{ 0\}). Define X_i as 2^i\cdot\mathbb 1_{\{i\} } , that is, ... 1 This answer is if the OP meant 'set'. The x values are all independent. For any value x, the probability that x is in both A and B is w^2. The probability that they don't clash at x is 1-w^2. So the probability that A and B clash nowhere would be (1-w^2)^\infty=0. Likewise, the probability they are equal is (w^2+(1-w)^2)^\infty=0 On ... 0 Keeping the definiton of the events A and B and applying Bayes' Theorem twice$$ P(A | B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B|A)P(A)}{P(B)}. $$I am answering the question: is it possible to have P(A|B) > P(A) ? From the equation above, that is indeed possible if$$ P(B|A) > P(B) $$or, in terms of the original random variables, if$$ P(X_2 ...

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Hint: For $r\in(0,1)$ $$P\left(R<r\right)=\int_{0}^{\infty}P\left(R<r\mid X_{1}=x\right)f_{X_{1}}\left(x\right)dx$$

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Definition(Measurable function): Let $f$ be a function from a measurable space $(\Omega , \mathcal F)$ into the real numbers. We say that the function $f$ is measurable if for each Borel set $B \in \mathcal B$ , the set $\{\omega; f(\omega) \in B\} ∈ \mathcal F$. Definition( random variable): A random variable $X$ is a measurable function from a probability ...

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In Bayesian inference, you've calculated the posterior distribution of $X$ using a uniform prior, which may or may not lead to a proper posterior distribution...

2

Here's a different argument. Define $$X_t = \int_0^t f(s,\omega) dW_s \\ Y_t = X_t^4 \\ Z_t = X_t^2.$$ Apply the Ito formula to $Y_t$ and $Z_t$: $$Y_t = 4 \int_0^t f(s,\omega) X_s^3 dW_s + 6 \int_0^t f(s,\omega)^2 X_s^2 ds \\ Z_t = 2 \int_0^t f(s,\omega) X_s dW_s + \int_0^t f(s,\omega)^2 ds$$ Now substitute and take expectations, thereby canceling the ...

0

The likelihood is simply the probability of observing the data under given parametric assumptions. Here: $P(x\in [a,b])=\frac{1}{b-a} \implies \mathcal{L}(a,b;n)=\frac{\prod\limits_{i=1}^n \mathbf{1}_{[a,b]}(x_i)}{(b-a)^n}$, the key to this is the numerator..most people forget this and then wonder why we don't set $a=b$. Thus, to maximize the likelihood, you ...

0

This question was posed by me on the final exam in Statistics 530 at the University of Pennsylvania. The problem was handed out only very shortly before it was posted here. I love the Stack Exchange, and I am sad to see that someone has abused it in this way. J. Michael Steele.

1

If you multiplied the integrand by $f_Y(y)$ then you would get the overall probability/density $f_X(x)$. However without that I'm not sure if there's any interesting meaning to the integral. If you integrate with respect to $y$ without taking the distribution of $y$ into account then in some sense you could probably get anything.

0

Maybe you are looking for a statement like this: The higher the probability of high values of $X$, the more likely it is that the firm defaults. That is, you are comparing probability distributions of $X$. Mathematically, you could use the notion of first order stochastic dominance (FOSD). Suppose you have two different distributions for $X$. Denote the ...

1

$A_n$ is increasing in $n$ because the event $$\bigg\{ \delta \sum_{i=1}^{n-1} X_i > b \bigg\}$$ is a subset of the event $$\bigg\{ \delta \sum_{i=1}^n X_i > b \bigg\},$$ simply because $X_n$ is nonnegative.

1

Usually we consider real-valued random variables, but there's no problem at all in allowing $\mathcal M$-valued ones, especially if if $\mathcal M$ is a discrete space. We could, if we wanted to be pedantic, encode messages by real numbers. EDIT: For a more general definition of random variables, see e.g. Wikipedia.

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Before answering the question, it seems necessary to dispel some misconceptions: There is no such thing as "the" transition matrix of a continuous time Markov process but a semi-group $(P_t)$ such that each $P_t$ is the transition matrix from $X_s$ to $X_{s+t}$, for every $s$. The matrix $T$ is not a transition matrix since it has some negative entries. ...

1

The fact that the thief can only rob one bank a day becomes irrelevant when we consider large numbers of trials, as everything becomes probabilistic anyway. We can thus treat the probability that the money will be safe in bank $k$ between days $n_{1}$ and $n_{2}$ as: $$\mathbb{P}[\text{money safe in bank }k \text{ between }n_{1} \text{ and ... 0 If t is any non-negative number, then consider n such that n\geqslant t and notice that$$\left\Vert \pi_{t}(\psi_1)-\pi_{t}(\psi_2)\right\Vert \leqslant \sup_{s\in\left[0,n\right]}\left\Vert \psi_1(s)-\psi_2(s)\right\Vert\leqslant 2^n\underbrace{2^{-n}\left\Vert \psi_1(s)-\psi_2(s)\right\Vert}_{\leqslant d(\psi_1,\psi_2)},$$hence$$\left\Vert ...

0

According to the red-sauce theorem, the possibility of you becoming green before seventy years of age is just as low as you becoming purple. I hope this is thoughtful enough.

1

The expectation $\mathbb{E}[Y_n]$ is equal to 1 for all $n$; the probability $\mathbb{P}\{Y_n > 0\} = \frac{1}{2^n}$ goes to $0$. However, note that with probability $1/2^n$ (i.e., when $Y_n$ is not $0$), $Y_n = 2^n$, which itself goes to infinity. The expectation is fixed and equal to $1$, the probability $p_n$ to be non-zero becomes smaller and ...

0

In the case of a Pascal distribution, also kwown as negative binomial random variable, the p.g.f. is: $$g(z)=\sum_{k=n}^{\infty}z^{k}\binom{k−1}{n−1}(1−p)^{k−n}p^n$$ Now, it's just a matter of massaging the summation in order to get a working formula. We start by effectively multiplying the summands by $1$, and thereby not changing the overall sum: ...

0

The CDF of $W$ is \begin{align*}F_W(w)&=P(W\le w)=P(\max\{X_1,X_2,\ldots X_8\}\le w)\\[0.2cm]&= P(X_1\le w, X_2\le w, \ldots, X_8 \le w)=^{\text{independent}}\\[0.2cm]&=P(X_1\le w)P(X_2\le w)\cdot\ldots\cdot P(X_8\le w)=^{\text{identically distributed}}\\[0.2cm]&=P(X_1\le w)^8=F_X(w)^8\end{align*} Since $W>0$ we have that ...

2

No, this is not true. We construct a counterexample in the plane. To make computations easy, we restrict ourselves to the unit square $U=[0,1]\times[0,1]$, but you will see that the idea extends to the entire plane easily. Define the set $N=[1/3,2/3]\times[1/3,2/3]$. Consider the probability density function $f(x,y)=\begin{cases} 0&(x,y)\not\in U,\\ ... 0 By assumption,$X=\alpha-Y $almost surely. Since$X$and$Y$are independent, so are$X$and$\alpha-Y$by properties of independent random variables. It follows that$X$is independent with itself, which implies that $$P(X\leq x)=P(X\leq x,\ X\leq x)=[P(X\leq x)]^2,\quad x\in{\Bbb R}.$$ Therefore,$P(X\leq x)=0$or$1$. Using the properties of ... 1 Let$f: [0,1] \to \mathbb{R}$be a continuous function. Since$[0,1]$is compact,$f$is uniformly continuous, i.e. we can choose$n \in \mathbb{N}$such that $$|f(s)-f(t)| < \frac{\varepsilon}{2} \quad \text{for all |s-t| \leq \frac{1}{n}.}$$ If we se set$t_j := j/n$,$j=0,\ldots,n, then \begin{align*} \mathbb{P} \left( \sup_{0 \leq t \leq 1} ... 1 Axiom of choice is irrelevant to the question of the existence of a non trivial measure on power set of reals. I suggest you start by reading Fremlin's "Real-valued measurable cardinals" in Set Theory of the Reals, H. Judah ed., Israel Math. Conf. Proceedings (1993), 151-305 2 Hint: Set Y_k = B_{t_k + s} - B_{t_{k-1}+s} where t_0 = 0. Note that X_k = Y_1 + \dots + Y_k, so you can express f(X_1, \dots, X_n) as a Borel function of Y_1, \dots, Y_n. Now observe that \{\mathcal{F}_s, \sigma(Y_1), \dots, \sigma(Y_n)\} are mutually independent \sigma-fields. Use a Dynkin lemma argument to conclude that \mathcal{F}_s ... 0 Hint: Consider the class \mathcal H of measurable bounded functions H such thatE((X-E(X\mid Y))\cdot H(Y,Z))=0.$$By hypothesis, \mathcal H contains every measurable bounded function H:(y,z)\mapsto h(y). On the other hand, the desired conclusion holds if \mathcal H contains every measurable bounded function H:(y,z)\mapsto H(y,z). Hence, the ... 2 Due to the Jensen's Inequality, for a convex function f and positive numbers \alpha_i with unit summation, we have:$$f(\alpha_1x_1+\alpha_2x_2+\ldots+\alpha_nx_n)\leq\alpha_1f(x_1)+\alpha_2f(x_2)+\ldots\alpha_nf(x_n)$$with equality if and only if f is linear or x_1=x_2=\ldots=x_n The important thing that you must pay attention to, is that there ... 2 A simple approach (which avoids the task of checking the nonnegativity of the function in your question) is to compare the hitting times \tau_1 and \tau_2 with the hitting time \theta of \pm2\pi by the process$$dY=\mathrm dW-\mathrm{sgn}(Y)\,\mathrm dt,$$starting from some |y|\leqslant2\pi. Since |\sin|\leqslant1, \tau_1 and \tau_2 are ... 1 Well, from a probabilistic point of view, you might not be able to have countable additivity at all. Since it might be that \Bbb R is the countable union of countable sets, then the only \sigma-additive measure which gives 0 to singletons, is the trivial measure. In that case you can still do some measure theory or probability, but you need to reduce ... 1 You are certainly not alone in asking this question to yourself: at least that seemed very strange and non-intuitive to me when I came across it. More to say, this definition of conditional probability is the Doob's version of the one proposed by Kolmogorov. None of them were immediately happily accepted by the mathematics community at the time they were ... 1 Let X_t be the number of busy servers in a M/M/\infty queuing system. You want to compute E(X_t|X_0=n). This can be done by considering the sum$$X_t=Y_t+Z_t,$$where \{Y_t|X_0=n\}\sim Binomial(n,e^{-\mu t}) is the number of servers that did not complete service in the interval (0,t] (out of the n that were busy), and Z_t\sim ... 5 The maximum M_n of interest is such that$$(1-2/\sqrt{n})K_n\leqslant M_n\leqslant1,\quad\text{where}\quad K_n=\max\{X_i\mid 1\leqslant i\leqslant\sqrt{n}\}.By independence, P(K_n\leqslant x)=P(X_1\leqslant x)^{\sqrt{n}}=x^{\sqrt{n}}\to0 when n\to\infty, for every x in (0,1). Hence K_n\to1 in probability, (1-2/\sqrt{n})K_n\to1 in ... 3 I have some partial results; perhaps someone can continue/correct my work. \begin{align*} &\phantom{{}={}}P\left(1-\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}>\epsilon\right)\\ &=P\left(1-\epsilon>\max_{1 \le i \le n/2} \left\{\left(1-\frac{2i}{n}\right)X_i\right\}\right)\\ &=\prod_{i=1}^{n/2} P\left(1-\epsilon ... 2 If we replace uniforms on (0,1) by uniforms on (0,w), the resulting random variable Y has the same distribution as wX, where X has Irwin-Hall distribution. In particular, \Pr(Y\le y)=\Pr(wX\le y)=\Pr(X\le \frac{y}{w}). It follows that if f_X is the density function of the Irwin-Hall, then Y has density f_Y(y)=\frac{1}{w}f_X(y/w). In a ... 0 What is true in general is that E[X1_F] = E[E[X |F]1_F] = P(F) E[X |F] $$but if F is not independent of X there is no equality in general. 0 Suppose X is the number of heads you get when you toss a coin (thus either 0 or 1) and F is the event that you get a head. Then \mathbb{E}\left[ X \mathbb{1}_{\{F\}} \right] is 1/2 and \mathbb{E} [X]P(F) is 1/4. 0 This is not true in general. For instance, for X = \mathbf{1}_F we have$$ \mathbb{E}[X \mathbf{1}_F] = \mathbb{E}[\mathbf{1}_F^2] = \mathbb{E}[\mathbf{1}_F] = \mathbb{P}(F) \neq \mathbb{P}(F)^2 = \mathbb{E}[\mathbf{1}_F] \mathbb{P}(F) = \mathbb{E}[X] \mathbb{P}(F)$whenever the probability of$F$is outside$\{0,1\}$. In fact, when$X = \mathbf{1}_G$... 0 The result is true, is given as e.g. Theorem I.50 of Protter's book "Stochastic integration and differential equations", and your argument is correct. In detail, to obtain the result, assume that$M$is a local martingale. Let$(T_n)$be a localising sequence, such that$M^{T_n}$is a martingale for each$n$. Then$M^{T_n\land n}$is a uniformly integrable ... 0 If$T$is a general stopping time, then no, as you can select e.g.$T=\infty$and the family you claim to be uniformly integrable is$(M_{\tau_n})$for some increasing sequence of stopping times$(\tau_n)$. Choosing$\tau_n = n$would yield that$(M_n)_{n\ge1}$is uniformly integrable for all martingales$M$, which is absurd. If$T\$ is a bounded stopping ...

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