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## Hot answers tagged probability-theory

7

There is a Borel set $E$ in $\mathbb R^2$ such that $F := \{x-y\colon (x,y) \in E\}$ is not a Borel set. Let $A := \{f \in \mathbf{C}\colon (f(1), f(0)) \in E\}$. Then $A \in \mathcal{B}_{\left[0,\infty\right)}$. How about $T(A)$? In fact $$T(A) = \{g \in \mathbf{C}\colon g(0)=0, g(1) \in F\}$$ and is not Borel. added Mar 10 Why is $T(A)$ not Borel? ...

3

Note that, for every $x\gt0$, $$X\geqslant x\mathbf 1_{X\geqslant x}.$$ (This is the step where one uses that $X\geqslant0$ almost surely.) It follows that $$E(X)\geqslant xP(X\geqslant x).$$ If $E(X)=0$ this inequality implies that $P(X\geqslant x)=0$. Finally, $$[X\gt0]=\bigcup_{n\geqslant1}[X\geqslant1/n],$$ hence, if every event in the RHS has ...

3

First of all, we choose $M \in \mathbb{N}$ sufficiently large such that $\sum_{m \geq M} 2^{-m} < \delta/3$. Moreover, we note that \begin{align*} \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \right) &\leq \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot 1_A \right)+ \mathbb{E}\left( 1 \wedge \sup_{s \leq m} |X_s^n-X_s| \cdot ... 3 On the same \Omega, try X uniform on \{0,1\} and Y=1-X, then \{X\ne Y\}=\Omega. Edit: Recall that in the probabilistic jargon, a random variable is just a measurable function, here X:\Omega\to\{0,1\} and Y:\Omega\to\{0,1\}, that is, for every \omega in \Omega, X(\omega)=0 or X(\omega)=1 and Y(\omega)=0 or Y(\omega)=1. A notation is ... 3 There isn't, because it doesn't. You are taking three uniformly distributed independent random variables X,Y,Z and forming a weighted average of Y and Z using X as a weight. The result of averaging will fall near 1/2 more often than near 0 or 1. I took 10^6 samples for illustration: And here is a mathematical derivation of the ... 3 Solving problems does take time, but it is usually time well-spent. You can learn a fair amount by watching the author develop the material, and by filling in details in the text. But, you learn far more by doing as many exercises as you can. Exercises develop technique, knowledge of and facility with the theorems and definitions, and give you the ... 2 You seem to be working towards a proof based purely on Markov theory. However, in this particular case, a simpler and more natural proof can be obtained through stochastic calculus, if you are willing to take a few theorems for granted. We proceed as follows. As you write yourself, let B denote a d-dimensional standard Brownian motion. Let \|\cdot\|_2 ... 2 Consider the function u defined on x\gt0 byu:x\mapsto k\log x-x^2+x.$$Then$$u'(x)=(k/x)-2x+1,\qquad u''(x)=-(k/x^2)-2.$$If k\gt0, u''\lt0 hence u' decreases from u'(0+)=+\infty to u'(+\infty)=-\infty, that is, u is increasing then decreasing, in particular u is uniformly upper bounded, that is,$$u(x)\leqslant c$$for every ... 2 This is saying that the average number of coin flips if you flip n fair coins each with heads probability p is np (you are calculating the mean of a Binomial(n,p) distribution). You can use linearity of expectation - if Y = \sum_i X_i where X_i is Bernoulli(p) and there are n such X_i, then Y is Binomial(n,p) and has that distribution. ... 2 You don't need to use induction or take derivatives; just note that k\binom{n\vphantom{1}}{k}=n\binom{n-1}{k-1}$$ \begin{align} \sum_{k=0}^nk\binom{n}{k}p^k(1-p)^{n-k} &=\sum_{k=0}^nn\binom{n-1}{k-1}p^k(1-p)^{n-k}\\ &=np\sum_{k=0}^n\binom{n-1}{k-1}p^{k-1}(1-p)^{n-k}\\ &=np\,\Big(p+(1-p)\Big)^{n-1}\\[12pt] &=np \end{align} $$2 Exercise: Let H:\mathbb R\to\mathbb R denote a non-decreasing function. Show that H is continuous from the right at a if and only if H(x_n)\to H(a) for at least one sequence (x_n) such that x_n\geqslant a for every n and x_n\to a when n\to\infty, if and only if H(a+1/n)\to H(a) when n\to\infty. Recall that H being continuous from ... 2 The question has been answered, so we give a different derivation. If p\ne 0, our series clearly converges, so the expectation exists. Call it a. On the first trial, either we have a success, in which case the expectation is 0, or we have a failure. In that case, we have wasted a trial, and the expectation is 1+a. It follows that$$a=(1-p)(1+a).$$... 2 Hint: You have confused throughout your solution (which however is methodologically correct) the roles of X and Y. You know that P(Y|X)=0.98\, (=1-False Negative) and P(Y|X^c)=0.03 Accordingly you have confused P(Y|X) with the actual required probability which is P(X|Y). For the calculation of P(Y) which will come in the denominator of ... 2 Counterexample Let$$S_n := \sum_{j=1}^n Y_j, \qquad n \in \mathbb{N} \tag{1}$$a simple random walk on \mathbb{Z}, i.e. Y_j \sim \frac{1}{2} (\delta_1+\delta_{-1}) independent identically distributed random variables. By Stirling's formula, we have$$\mathbb{P}(S_{2n}=0) = 2^{-2n} {2n \choose n} \sim \frac{1}{\sqrt{\pi n}} \qquad \text{and} \qquad ...

2

Recall that the conditional expectation ${\rm E}[X\mid Y]$ has the property that $$\int_A {\rm E}[X\mid Y]\,\mathrm dP=\int_A X\,\mathrm dP,\quad A\in\sigma(Y).$$ Since $\{Y=y\}\in \sigma(Y)$ for all $y$, we have $$\int_{\{Y=y\}}{\rm E}[X\mid Y]\,\mathrm dP=\int_{\{Y=y\}}X\,\mathrm dP=\int_\Omega X\mathbf{1}_{\{Y=y\}}\,\mathrm dP.$$ At last, recall that ...

2

The conditional distribution of $N$ given that $K = k$ is a Poisson distribution with parameter $q\lambda$ but displaced $k$ to the right; that is, (conditionally) $N$ is of the form $M+k$ where $M$ is Poisson$(q\lambda)$ and so $$E[N\mid K = k] = E[M+k] = E[M]+k = q\lambda + k.$$ Or, without spending time thinking about the matter, write $m = n-k$ and the ...

2

This is a kind of "uniform ergodic theorem" and extends naturally the case $X_k=X$ for each $k$. Notice that $X_k=X_k-X+X$ and by Birkhoff's ergodic theorem, $$\frac 1n\sum_{k=0}^{n-1}X\circ T^k\to \mathbb E[X\mid\mathcal I]\quad\mbox{ a.s.},$$ where $\mathcal I$ denotes the $\sigma$-algebra of invariant sets, that is, $\mathcal I=\{A, T^{-1}(A)=A\}$. If ...

2

Hint: We can say :$$\omega \in C \Leftrightarrow (\forall m \in \Bbb N )(\exists p \in \Bbb N)( \forall k \geq p) \quad \left|\sum_{i=k+1}^{+\infty} X_k(\omega)\right| < \frac 1{m+1}$$ That gives:$$C=\bigcap_{m \in \Bbb N}\bigcup_{p \in \Bbb N}\bigcap_{k \geq p} Y_{m,p,k}$$ Where : $$Y_{m,p,k}=\left\{\omega \in \Omega / \left| \sum_{i=k+1}^{+\infty} ... 2 I assume drawing is done without replacement. The probability that the values are equal is \frac{3}{51}. This is because whatever card we draw first, the probability of matching it is \frac{3}{51}. So the probability they are not equal is \frac{48}{51}. Thus by symmetry the probability the second is higher than the first is \frac{24}{51}. ... 2 In order to make the inner product well-defined, we talk about L^2(\Omega,\mathcal F,\mu), where (\Omega,\mathcal F,\mu) is the underlying probability space. But we then extend condition expectation to integrable random variables. We use a projection over the closed subspace L^2(\Omega,\mathcal N,\mu), that is, the vector subspace which consists of ... 2 No. Counterexample:$$ X=0; Y = \text{anything not }G\text{ mesurable} $$But if it is true for any bounded r.v. this is correct (this is the definition of the conditionnal expectation). Yes, look at the definition. E[X|G] is the only G mesurable r.v. such as for every Gmesurable bounded r.v. Y,$$ E[XY] = E[E[X|G]Y] $$2 If X_n\to 0 in distribution, then any \varepsilon does the job. The converse is harder. Here it's the proof of Levy's continuity theorem which will be used. Denoting by \varphi_n the characteristic function of X_n and \mu_n its distribution, we indeed have the equality ... 2 Let (X_t)_{t \geq 0} be a non-negative solution of the SDE$$X_t - x = 3t + 2 \int_0^t \sqrt{X_s} \, dB_s \tag{1}$$for x \geq 0. Applying Itô's formula to f(y) = \frac{1}{\sqrt{y}}, we find$$\frac{1}{\sqrt{X_t}} - \frac{1}{\sqrt{x}} = - \int_0^t \frac{1}{X_s} \, dB_s.$$For \tau_{a,b} := \inf\{t \geq 0; X_t \notin (a,b)\}, 0<a<b, this ... 2 We need to check that the random variable X(t) is measurable with respect to \sigma-algebra \mathcal F_t for each t\ge0. If t\in[0,1], the only value that X(t) can take is 0. So we need to find the smallest \sigma-algebra that contains X^{-1}(t)(\{0\})=\Omega. Such \sigma-algebra is \{\emptyset,\Omega\}. If t\in(1,2], the range of ... 1 Let p be the chance that A wins. Then B wins 1-p. A can win by throwing 10, or by throwing something else, having B not throw a 9, and then winning from start. As the chance of a 10 is \frac 3{36} and a 9 is \frac 4{36}, we have p=\frac 3{36}+(1-\frac 3{36})(1-\frac 4{36})p 1 Consider the following two scenarios: 1) You have two independent Poisson processes N_1(t) and N_2(t) with rates \lambda p and \lambda (1-p). N(t) = N_1(t) + N_2(t) is their sum. This is a Poisson process with rate \lambda. 2) You have a Poisson process N(t) with rate \lambda. Each time an event occurs, a coin is tossed: with ... 1 Hint: We get the independence result if we can show that$$\mathbb{E}\left[\mathrm e^{\mathrm i\left(aN_1(s)+bN_2(t)\right)} \right] = \mathbb{E}\left[\mathrm e^{\mathrm i aN_1(s)} \right]\mathbb{E}\left[\mathrm e^{\mathrm i bN_2(t)} \right], $$for all s\leq t and a,b. Once the case s=t is done, the case s<t is formal, based on conditioning and ... 1 Set W_t := A \cdot B_t where A is an orthogonal matrix. Since W_0=A \cdot x we see that (W_t)_{t \geq 0} starts at A \cdot x. As t \mapsto B_t is continuous, we find that t \mapsto A \cdot B_t is continuous. Let 0 \leq s \leq t. Then,$$\begin{align*} \mathbb{E}e^{\imath \, \xi^T \cdot (W_t-W_s)} = \mathbb{E}e^{\imath \, \xi^T (A \cdot ...

1

Let $n$ be the number of questions. We will calculate the chance that exactly four questions are repeated out of eight. There are ${n-8 \choose 4}{8 \choose 4}$ ways to choose four matching and four non-matching, and $n \choose 8$ ways to choose the questions overall. So we want $$\frac {{n-8 \choose 4}{8 \choose 4}}{n \choose 8}=0.05$$ I find $36$ gives ...

1

If there are $n$ question is the bank and we have already picked our 8 for the first time, then the $n$ questions are split into two categories: the 8 chosen the first time and the remaining $n-8$ not chosen the first time. Thus the question can be restated as: what is the smallest $n$ for which  ...

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