# Tag Info

0

Why do you think the expression implies such a lack of independence? Intuitively, the Markov property allows expectations conditional on randomness at a single point in time, $t$, to be replaced by expectations conditional on all of the randomness at points in time up to and including time $t$. Thats why $\mathfrak{F}_t$, which includes all the ...

0

It is not difficult to see that the process $$W_t := \begin{cases} t B_{1/t}, & t > 0, \\ 0, & t=0 \end{cases}$$ defines a Brownian motion and $$\mathcal{G}_t := \sigma(B_u; u \geq t) = \sigma\left(W_u; u \leq \frac{1}{t} \right) =: \mathcal{H}_{\tilde{t}}$$ for ${\tilde{t}} := 1/t$ is the canonical filtration of $(W_t)_{t \geq 0}$. Moreover, ...

0

$P(X_1>0)>0$ implies $\zeta<1$. If $P(X_1>0)=0$, then $\zeta=1$ giving you a divergent upper bound when summing $F_n(t)$. In your integration, you are assuming $X_1$ has continuous density. On the other hand if $X_1$ has a jump, in particular if $X_1=0$ with probability 1,then the expectation of $e^{-X_1}$ is clearly 1 as well. You've written ...

0

Hint: Use Doob's maximal inequality $$\mathbb{P} \left( \sup_{s \leq t} |M(s)| \geq \delta \right) \leq \frac{1}{\delta^p} \mathbb{E}(|M_t|^p), \qquad \delta>0,$$ which holds for any continuous martingale $(M_t)_{t \geq 0}$ such that $M_t \in L^p$, $p \geq 1$.

2

No: let $\Omega=\{a,b\}$, $\mathcal{F}=2^\Omega$ and suppose that your filtration is $\mathcal{F}_1=\{\emptyset,\Omega\}$ and, for any $k\ge 2$, $\mathcal{F}_k=2^\Omega$. Let $\mathbb{P}$ be given by $\mathbb{P}(\{a\})=\mathbb{P}(\{b\})=\frac{1}{2}$. Let $X_1\equiv 1$ and for $k\ge 2$ $X_k(a)=2$, $X_k(b)=0$. Then $(X_k)$ is a nonnegative converging ...

0

This algorithm is attributed to John Von Neuman, and works as follows. We know that $\mathrm{P(Head)} = q \neq \frac{1}{2}$. Now, let's enumerate the probabilities for the set of 2 flips: $$\mathrm{P}(HH) = qq\\ \mathrm{P}(HT) = q(1-q)\\ \mathrm{P}(TH) = (1-q)q\\ \mathrm{P}(TT) = (1-q)(1-q)\\$$ Note that $P(TH) = P(HT)$! Given that Heads then Tails is ...

1

No, $T_n<\infty$ does not hold true. Proof: Suppose that $T_n<\infty$ almost surely. Since $(X_t)_{t \geq 0}$ is a martingale, $X_0 = 1$, the optional stopping theorem yields $$\mathbb{E}X_{T_n \wedge t} = 1$$ for any $t \geq 0$. Since $|X_{T_n \wedge t}| \leq n$ for all $t$, we get $$\mathbb{E}X_{T_n}=1 \tag{1}$$ by the dominated convergence ...

0

Hint: Let $X$ denote the sum of eyes by throwing two dice and let $E$ be some event. Then (*) $P\left(E\text{ and } X=4\right)=P\left(E\mid X=4\right)P\left(X=4\right)$ So finding $P\left(E\text{ and }X=4 \right)$ and $P\left(X=4\right)$ leads to finding $P\left(E\mid X=4\right)$. Apply this on the events mentioned in a), b) and c). Let $D_{1}$ be the ...

1

You have $3$ possible events in your probability-space: First Die | Second Die -----------|------------ 1 | 3 -----------|------------ 2 | 2 -----------|------------ 3 | 1 Out of these $3$ events: There is $1$ event in which the first die gave 3, hence the probability is $\frac13$ There are $2$ events in ...

0

I doubt $Z$ has a simple distribution. But it might be possible to describe its moments. For example if you have a random variable $X_i$ with a Poisson distribution with mean $\lambda_i$ then its variance is also $\lambda_i$ and its second moment is $\lambda_i+\lambda_i^2$. So if you have a random variable $Y_i$ which is $X_i$ with probability $p$ and $0$ ...

0

For two random variables $x_1$ and $x_2$ having means $u_1$ and $u_1$, and variances $v_1$ and $v_1$ if $z= a \cdot x_1 + b \cdot x_2$ then, $\operatorname{mean}(z) = a\cdot u_1 + b\cdot u_2$ and $\operatorname{variance}(z) = a^2 \cdot v_1 + b^2 \cdot v_2$ Hence, for your problem, $\operatorname{mean}(Z) = 0$. $\operatorname{variance}(Z) = (3/4)^(2n-2) + ... 0 Hints: Recognise that the only source of uncertainty in this model comes from the Wiener-process term, $$\dfrac{\mathrm{d}S_t}{S_t}{}={}\mu_{t}\mathrm{d}t{}+{}\color{red} {\sigma\mathrm{d}W_{t}}\,.$$ Consequently, thinking about what happens to the solution,$C\left(t, S_{t}\right)$, as$\sigma\to 0$, should give you the insight you seek. So, ($i$) ... 0 Use the tower property to show that for any$k \ge n$, we have $$E[X_{k+1} \mid \mathcal{F}_n] \ge E[X_k \mid \mathcal{F}_n].$$ So we may rewrite$Y_n$as $$Y_n = \lim_{k \to \infty} E[X_k \mid \mathcal{F}_n] \tag{*}$$ since the limit of an increasing sequence equals its supremum. Now$E[Y_{n+1} \mid \mathcal{F}_n] = Y_n$follows from the conditional ... 2 Let$f(x) := x^a$for some fixed$a>0. Then $$f'(x) = a x^{a-1} \qquad f''(x) = a (a-1) x^{a-2}.$$ Since by Itô's formula $$f(X_t)-f(X_0) = \int_0^t f'(X_s) \, dX_s+ \frac{1}{2} \int_0^t f''(X_s) \, d\langle X \rangle_s$$ we get \begin{align*} Y_t - Y_0 &= f(X_t)-f(X_0) \\ &= a \int_0^t X_s^{a-1} \,d X_s + \frac{1}{2} a (a-1) \int_0^t ... 0 A stochastic process is an ensemble of deterministic waveforms, or realizations, where each waveform is a function of time. Just as the random variable X maps each outcome \omega in sample space \mathcal{S} to \mathbb{R}, the random process X(t) maps each outcome to a deterministic function of time. If time is fixed, say, at t_1, the random ... 3 Recall the following statement from probability theory: Theorem: Let (Y_n)_{n \in \mathbb{N}} be a sequence of pairwise independent random variables. Then Y:=\limsup_{n \to \infty} Y_n is constant almost surely. Fix t \geq 0 and a sequence (t_n)_{n \in \mathbb{N}}, t_n \neq t, such that t_n \to t. By the continuity of the process (X_s)_{s ... 2 First of all, there are several typos in your calculations (e.g. it should read \int_0^t W_s^2 \,ds instead of \int_0^t W_t^2 \, ds). Your calculation goes wrong when you write\mathbb{E} \left( \int_0^t W_s^3 \, dW_s \right) = \frac{\mathbb{E}(W_t^4)}{4} - \frac{3}{2} \int_0^t V(W_s) \, ds = \frac{\color{red}{3t^4}}{4} - \frac{3t^2}{4}.$$(I don't ... 1 we know the mean E[\epsilon_j]=0 so the variance E[\epsilon_j^2] - (E[\epsilon_j])^2 is E[\epsilon_j^2] which is 1 due to independence var(Z_i)=\sum_{j=1}^i var(\epsilon_j)=\sum_{j=1}^i 1 = i 1 Ito's Lemma: For suitable stochastic process X_t(t, B_t),$$ dX_t{}={}\left(\dfrac{\partial}{\partial t}X_t{}+{}\mu_b\dfrac{\partial}{\partial b}X_t{}+{}\dfrac{1}{2}\sigma_b^2\dfrac{\partial^2}{{\partial b}^2}X_t\right)dt{}+{}\sigma_b\dfrac{\partial}{\partial b}X_tdB_t\,. $$where \mu_b{}={}0 and \sigma_b{}={}1 for brownian motion, B_t . From ... 1 Lemma: On the space \tilde{\Omega} := \{f: [0,\infty) \times \Omega \to \mathbb{R}; f càdlàg\} we define a norm by$$\|f\|_{L^1} := \mathbb{E}(\|f\|_{\infty}) := \mathbb{E} \left( \sup_{t \geq 0} |f(t)| \right).$$Then$$L^1(\Omega; D[0,\infty)) := \{f \in \tilde{\Omega}; \|f\|_{L^1}<\infty\}$$is a complete normed space. Proof: As a composition of ... 1 Since Y is a positive random variable, only outcomes with w=x+y>x=z are possible. Because all positive values are possibilities for Y and X, the pdf will be posive for all z=x>0 and all w=x+y>x=z. It follows that pdf is positive at {(z, w): z>0, w>z}. No need to compute the pdf to determine where pdf is positive in above reasoning. In fact, the above ... 0 Assuming k\in\mathbb{N},$$ \begin{eqnarray*} \mathbb{E}\left[X_n X_{n+k}\right]&{}={}&\mathbb{E}\bigg[\varepsilon_n\left(\varepsilon_n-\varepsilon_{n-1}\right)\varepsilon_{n+k}\left(\varepsilon_{n+k}-\varepsilon_{n+k-1}\right)\bigg]\newline ... 1 Otherwise stated\mathcal{A}_{\tau}$is formed of events that can be defined by questions that can be answered (positively or negatively) only if you know whether$\tau$as happened at time any time$n$. Notice that involving the knowledge of$X$in the informal definition is not necessary IMO unless the filtration is the natural filtration generated by ... 0 Please look here: http://en.wikipedia.org/wiki/Perron%E2%80%93Frobenius_theorem. You should only to verify that your matrix P satisfies the conditions of this theorem but it shouldn't be so difficult. 0 Locally Compact spaces are not necessarily metric, and not necessarily separable then not a Polish space which is the natural setting for probability and the state space of Markov process. The sate space can be a functionnal space see here. Definition of trajectories need some elementary notation.$(\Omega, \mathscr{F}, \mathscr{F}_t, \mathbb P)$is a ... 1 See here for the fact that$\int_0^t B_u du$follows the normal distribution with zero mean and variance$\dfrac{t^3}{3}$. See here for moment generating function(MGF) of normal distribution 1 First note that sets of the form$\{t\in\mathbf{R}_{+}, B_t = x\}$for$x\in\mathbf{R}$are non empty by limit properties of the brownian motion. I assume more generally that$B$starts at$x$, not$0$, but that$a < x < b$. (I may have switch your$a,b$with mine, but who cares.) Let us first recall three "optional sampling" results (théorème d'arrêt ... 1 Hint: By the optional stopping theorem, both $$(B_{t \wedge T_{a,b}})_{t \geq 0} \tag{1}$$ and $$(B_{t \wedge T_{a,b}}^2- (t \wedge T_{a,b}))_{t \geq 0} \tag{2}$$ are martingales. Use the dominated convergence theorem to conclude from (2) that$T_{a,b} \in L^1$. Then let$t \to \infty$to get $$\mathbb{E}T_{a,b} = \mathbb{E}(B_{T_{a,b}}^2) \qquad ... 0 Yeah, thanks. It was quite simple. My solution was to verify the martingality of y(t) following the three conditions:1 – Measurability Condition. y(t) \in Ft natural filtration since y(t) is a transformation of Bt which is a Ft-martingale for the probabilistic properties of Brownian Motion. 2 – Integrability Condition,$$ E |e^{2B_t - \alpha t}| < ... 1 $$Y_t = \mathrm{e}^{2B_t-\alpha t}$$ using Ito $$dY_t = -\alpha Y_t dt + 2Y_tdB_t + \frac{4}{2}Y_t dB_t^2 = -\alpha Y_t dt + 2Y_tdB_t + 2Y_t dt = (2-\alpha)Y_tdt + 2Y_tdB_t$$ thus martingale requires a driftless SDE thus$\alpha = 2$. Was this the way you performed it? 0 You get the value of the pay-off (assuming it is a european option of expiray$T$) by looking at$P_T$, as by definition, this is the pay-off. Indeed, for a european call of strike$K$and expiry$T$you have: $$P(t,S_t) = S_t \varphi( d_{+} ) - Ke^{-r(T-t) } \varphi( d_{-} )$$ where$\varphi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-\frac{t^2}{2}}dt$... 1 Here's a somewhat expanded answer, since this is a fairly straightforward question if you've covered the relevant definitions. Let$m < n$, then we have that$\mathbb{E}(X_n|I_m) = \mathbb{E}(\mathbb{E}(Y|I_n)|I_m)$. But from the definition of a filtration, we know that$I_m \subseteq I_n$, so the law of iterated expectations applies, and it follows ... 0 You can efficiently generate the number of successful trials via the Poisson-Binomial distribution. The PMF need only be calculated once for a given set of$P_i$, and variates generated by the method of your choice - using simple inverse sampling I generated >5 million / second on an old atom-based netbook. Once you have the number of successes (or a ... 0 Here's a suggestion: Lets say you have$n$trials ($T_i$), where$p_i = P(T_i=1), q_i=1-p_i$. Then there are$2^n$combinations of successful trials (e.g.,$(1,0,0,1)$is one such combination if$n=4$). Set an$2^n$-vector,$\mathbf{P}:=(0,0,...0)$Let$I:=\{0, 1,2,...2^n-1\}$be a set of integers For each$i\in I$, calculate its binary expansion as an ... 1 It is more intuitive to think$\mu(dx)$in the form of$p(x) dx$.$\mu$is a probability measure and generally it is assumed to be absolutely continuous with respect to Lebesgue measure$dx$. So let's put (informally)$\mu(dx) = p(x) dx$where$p(x)$is the probability density function of this probability measure and$dx$is the Lebesgue measure. Then ... 0 I see what you mean. The general rule is that you cannot find the KL eigenfunctions in closed form, just like you cannot find eigenvectors (for$D>5$) in the discrete case. This is possible in a few special cases only. So you have to make an approximation. Two suggestions: 1) Do it numerically. No matter how ugly is your kernel the transformation is ... 2 A filtration generated by Brownian motion simply means the smallest filtration with respect to which Brownian motion is adapted i.e. $$\mathcal{F}^B_t = \sigma \{B_s, \: s\leq t\}$$ On the other hand, a filtration for the Brownian motion,$\mathcal{F}_t$is one to which the Brownian motion is adapted AND we have that for all$s<t$,$B_t - B_s$is ... 2 The product topology on$\mathbf{R}^{[0,T]}$is the topology for which a sequence (or a net) in$\mathbf{R}^{[0,T]}$converges if and only if all its projections to the component spaces$\mathbf{R}$converges. This is the same as pointwise convergence of functions from$\mathbf{R}^{[0,T]}$. This topology is therefore called topology of pointwise convergence, ... 2 OK, as suggested by @TheBridge, I found that problem 4.2 on page 60 of Karatzas and Shreve's Brownian Motion and Stochastic Calculus (2nd edition) demonstrates my question. The answer is positive. To be more succinct, the Borel sigma-algebra on$C[0,t]$is the same as the product sigma-algebra$\mathbb{R}^{[0,t]}$(i.e., the smallest sigma-algebra such that ... 0 I have a feeling that there's a much simpler reasoning for this, but nevertheless, here's a counterexample. Let$F(x)=1-e^{-x}$and$G=F^{-1}$. Let$W_t$be a Brownian motion starting at zero on some probability space$(\Omega_1,\mathcal{F}_1,P_1)$. For$T=G(1/2)*2$denote$A=\{W_T-W_{T/2}>0\}\subset\mathcal{F}_1$. Define the random variable$\eta=I_A$, ... 1 Hint: It suffices to calculate expectations of the form $$\mathbb{E}(X_t \cdot X_s)$$ where$s \leq t$. Setting$X_t = W_t^2, we have \begin{align*} \mathbb{E}(X_t X_s) &= \mathbb{E}\big( ((W_t-W_s)+W_s)^2 W_s^2 \big) \\ &= \mathbb{E}((W_t-W_s)^2 W_s^2) + 2 \mathbb{E}((W_t-W_s) W_s^3) + \mathbb{E}(W_s^4). \end{align*} Now use thatW_s \sim ...

0

Have you tried using $R$? For instance see http://statisticalrecipes.blogspot.com/2013/01/easy-introduction-to-markov-chains-in-r.html For visualization see here for example. By the way, this visualization looks really nice, but runs in Javascript, not R.

1

Hint: The stochastic integral $$M_t := \int_0^t e^{-as} \, dB_s$$ is a martingale. Hence, $\mathbb{E}M_t = \mathbb{E}M_0=0$. In order to calculate the variance of $X_t$ use Itô's isometry. Alternative approach: Since stochastic integrals are martingales, we have $$\mathbb{E}X_t- \mathbb{E}X_0 = \int_0^t a \cdot \mathbb{E}X_s \, ds,$$ i.e. $m(t) := ... 1 I'm not sure what you mean by "justify the shape generated from those equations". An iterated function system, or IFS, is a non-empty list of functions$(f_i)_{i=1}^m$mapping a a complete metric space (often$\mathbb R^n$) to itself. If each of those functions is a contraction, then one can prove that there is a unique, non-empty compact set$E$with the ... 0 As I suggested in the question, diverging from the hint, one could also proceed as follows: Note that$\mathbb{P}(\limsup_{t\rightarrow\infty}\frac{\left|B_{t}\right|}{t}>\frac{1}{M})\leq\mathbb{P}(\bigl\{\sup_{t\in\left[n,n+1\right]}\frac{\left|B_{t}\right|}{n}>\frac{1}{M}\bigr\}\;\mbox{i.o.})=0$for all$M\in\mathbb{N}$by applying Doobs Inequality ... 1 Hence it suffices to show that$\max_{\left\lfloor t\right\rfloor \leq s\leq\left\lfloor t\right\rfloor +1}\left|\frac{B_{s}-B_{\left\lfloor t\right\rfloor +1}}{\left\lfloor t\right\rfloor }\right|$converges a.s. to$0$[when$t\to\infty$.] Here is a direct approach. For every$n\geqslant1$, let $$X_n=\max_{n \leqslant s\leqslant n ... 1 Suppose that we keep on playing even if we go bankrupt (possibly ending in negative numbers as if ending with debt), the money you have after n coin-flips would be distributed like X = K-n + 2 Bin(0.5,n) where Bin(p,n) denotes a binomial distribution. An important observation is that the probability of reaching 0 pounds somewhere during gambling and ... 0 It depends on the space on which you define A, but if we take for example C^{\infty}\left([0,\infty[\right), we can choose a basis of functions and see if we can find coefficients such that a linear combination of basis functions is an eigenfunction. This is actually really similar to the finite dimensional case. Concretely, you could use the Fourier ... 0 One trick to apply DCT or similar statements is following: Let's stay tou want to show \int X_t \rightarrow \int X as t \rightarrow a. If you show this for every sequence t_n \rightarrow a, then you are done. In this case you are reduced to the more familiar situation. You can also consider this by proof by contradiction. If such a statement didn't ... 0 Given \mathscr{F}_1, \beta_ t = W_{t + 1 } - W_1 is a standard Brownian Motion independent of W_1. So, for t \geq 1,$$ \mathbf{E}[ W_t^2 | \mathscr{F}_1 ] = \mathbf{E}[ (\beta_{t-1} + W_1 )^2 | \mathscr{F}_1 ] = t-1 + W_1^2. $$So that,$$ \mathbf{E}\left[ \int_1^2 W_r^2 dr \big | \mathscr{F}_1 \right] = 1/2 + W_1^2.$\$

Top 50 recent answers are included