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In addition to the hints, we need the following identities related to derangements: (1) $\displaystyle\sum_{i=0}^{n}\frac{!i}{i!~(n-i)!}=1,~(n\ge0),$ (2) $\displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)}{i!~(n-i)!}=1,~(n\ge1),$ (3) $\displaystyle\sum_{i=0}^{n}\frac{!i~(n-i)~(n-i-1)}{i!~(n-i)!}=1,~(n\ge2),$ where I have adopted the subfactorial notation. The ...

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The pdf of $X=(X_1,\,X_2)$ is $$f(x_1,x_2)=\begin{cases} \frac{1}{4} & \text{for } (x_1,x_2)\in Q=\{(-1,0), (1,0), (0,-1), (0,1)\}\\ 0 & \text{otherwise} \end{cases}$$ and the variable $X=(X_1,X_2)$ can be represented in tabular form $$\begin{pmatrix} (X_1,X_2)\\ f(x_1,x_2) \end{pmatrix}= \begin{pmatrix} (-1,0) & (1,0) & (0,-1) & ... 2 Check out George Lowther's blog: https://almostsure.wordpress.com/2010/01/03/the-stochastic-integral/ Especially Lemma 3, answers your question "why cadlag?" 2 The number of offspring of a cell is X. The probability that X=k is given by: ~\mathsf P(X=k) = qp^k for k\in\{0, 1, 2,\ldots \} and q=(1-p). The expected value of X is given by : ~\mathsf E(X) = \sum_{k=0}^\infty k~\mathsf P(X=k) Thence, using Geometric Series closed form (when \lvert p\rvert<1):$$\begin{align}\mathsf E(X) ~=~& ...

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I think you misunderstood the meaning there. They talked about "contingent" claims, these are random variables given some other process. A simple example will be something like this: you have a simple SDE, $dX = dB$, with $X_0 = a$, the (strong) solution is, of course, $X_t = a+ B_t$. Now, you have another process on the same Brownian Filtration: $dY ... 2 Below is an outline. The grey areas are spoilers — reveal them by placing your mouse over them if you want details, but I'd suggest you try by yourself before. Compute the expression of the CDF$F^{(n)}_X$of$X\stackrel{\rm def}{=} \max(X_1,\dots, X_n)$, using independence. Recall that for$t\in\mathbb{R}$, $$F^{(n)}_X(t) = \mathbb{P}\{ X \leq t \} = ... 2 If X has generator G and b>0 is a scalar, then X(bt) is a Markov process with generator bG. If G_1 and G_2 are generators on the same state space, and (writing D(G_i) for the domain of G_i) D(G_1)\cap D(G_2) is large enough, then G_1+G_2 is the generator of another Markov process. Under the appropriate conditions, the semigroup S ... 1 You're certain to take exactly one step each from 1 to 2 and from 3 to 4; that's the constant 2. At 2, you have a Bernoulli experiment with success probability \frac23 that you perform until it succeeds, and at 4 you have a similar Bernoulli experiment with success probability \frac35. The expected time until a Bernoulli experiment succeeds ... 1 Here is an example in discrete time explaining why the Markov property is crucial to guarantee that solutions do not cross, in the sense that if two solutions coincide at some given time then they coincide at any later time. Consider the AR2 process (x_n)_{n\geqslant0} defined by some initial conditions (x_0,x_1) and by the recursion ... 1 Hint: The density f(a)=\frac12e^{-|a|} belongs to the Laplace distribution. The Laplace distribution arises when you subtract two iid exponential(1) variables. Second hint: If U has uniform distribution on [0,1] then X:=-\ln (U) has exponential(1) distribution. 1 Just for your reference as an alternative solution, though more complex. Let X be the number of coin tosses until we get a success. Then$$ \Pr(X = i) = q^{i-1}p $$The corresponding generating function is$$ f(x) = \sum_{i=1}^\infty q^{i-1}p\cdot x^i = \frac{p}{q}\cdot\sum_{i=1}^\infty (qx)^i = \frac{p}{q} \cdot \frac{qx}{1-qx}=\frac{px}{1-qx} $$Now ... 1 \{Y=n+i\} is the event that among n+i-1 experiments there were i failures, and secondly also that the n+i-th experiment was a success. There are \binom{n+i-1}{i} ways of selecting i experiments out of n+i-1. Each of these selections goes along with a probability of p^{n-1}q^i of occurring. The n+i-th experiment has probability p to ... 1 Let \varphi(\theta)=\mathbb E[e^{\theta X_1}] be the moment generating function and \psi(\theta)=\log\varphi(\theta) the cumulant generating function of X_1. We compute$$ \varphi(\theta) = e^{-\theta}\mathbb P(X_1=-1)+e^{\theta}\mathbb P(X_1=1)=\frac12\left(e^{-\theta}+e^{\theta}\right)$$and$$\psi(\theta) = \log\left(\frac 12 ... 1 Express$Y$as an Ito process: $$dY_t=X_t\,dt = X_t\,dt +\ 0\,dB_t.$$ 1 Let's assume we are in the context of a filtered probability space$(\Omega,(\mathcal F_t),\Bbb P)$satisfying the usual conditions. Here are a few features of optionality that make it important. If$T$is a stopping time and$A\in \mathcal F_T$then there is a bounded optional process$Z$such that$Z_T=1_A$on$\{T<\infty\}$. If$Z$is an optional ... 1 I think you've got the terminology mixed up a little. You have properties of a process (such as being adapted to a filtrartion, being progressively measurable, or being predictable), and then you have stopping and optional times (where the difference is, roughly speaking, merely that optional processes cannot answer questions regarding the "current time"). ... 1 $$X = \sum_{i=1}^n X_i$$ where$X_i \in \{ -1, 0, 1 \}$is the change in$x$at step$i$. Since the variables$X_i$are independent and each has mean 0, this implies that $$E(X^2) = \sum_{i=1}^n E(X_i)^2 = n E(X_1^2)$$. From here, it should be easy to calculate$E(X_1^2)$explicitly. 1$X_1 X_2=0$so$E[X_1 X_2]=0$and thus, since$E[X_1]=E[X_2]=0$, the covariance and correlation must be zero. But, for example,$X_1=1 \implies X_2=0$so they are not independent 1 "Progressive" and "optional" are refinements of "adapted" involving some degree of joint measurability of$(\omega,t)\mapsto X_t(\omega)$. The three notions coalesce in discrete time. Consider, for example, the notion "optional" in a discrete-time setting. So let$(\Omega,(\mathcal F_n)_{n\ge 0},\mathcal F,\Bbb P)$be a filtered probability space. The ... 1 This might be helpful: Theorem 29 on page 173 of Protter's "Stochastic integration and differential equations": Let$M$be a local martingale and let$H\in \mathcal P$be locally bounded. then the stochastic integral$H\cdot M$is a local martingale. I'm reading that book, but haven't got that far yet, will be interested to see if someone can provide a ... 1 There are at least two reasons. I am certain of the second, only 60% of the first one, so please keep this in mind. 1. Look at equation (2.7);$F_t\$ is a stopped version of the integral of the progressively measurable process. It should follow that (at least restricted to locally bounded/integrable progressively measurable processes) that the integral ...

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I realize this is a late answer to this post, but it still makes the top two to three results on Google for "ensemble average" and an answer has not yet been officially accepted. For posterity, I figured I would try to answer it to the best of my ability in the way that the question has been phrased. First, it is important to have a broad understanding of ...

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