I observe a sample from a distribution that I expect to be the hitting time   $$\tau = \inf\{t&gt;0| X(t)&gt;a\}$$  where $X(t)$ is a Lévy process with $X(0)=0$ and $a$ is some constant. $X$ is not a Brownian motion and the experimental fit to the  is poor.   However, I do not need to know the exact formula for the law of $\tau$. For my needs I only need to know that the expectation of $\tau$ is infinite (as in the case of tau for a Brownian motion). Is it possible to formulate and test this as a statistical hypothesis?
66043
Is it possible to test for finiteness (or existence) of the variance of a random variable given a sample? As a null, either {the variance exists and is finite} or {the variance does not exist/is infinite} would be acceptable. Philosophically (and computationally), this seems very strange because there should be no difference between a population without finite variance, and one with a very very large variance (say > $10^{400}$), so I am not hopeful this problem can be solved.  One approach that had been suggested to me was via the Central Limit Theorem: assuming the samples are i.i.d., and the population has finite mean, one could check, somehow, whether the sample mean has the right standard error with increasing sample size. I'm not sure I believe this method would work, though. (In particular, I don't see how to make it into a proper test.)
28918
I proceeded by finding $\operatorname E(\bar X).$ I considered $\bar X$ as a constant and simply got the term itself. This should suggest that $\bar X$ is not an unbiased estimator of $\theta$. That's obvious too cuz it's a constant(from what I assumed). However, I'm also aware that $\mu = \theta$ so that's probably not right.  I have to demonstrate first that $\bar X$ is an unbiased estimator of $\mu$.
25365
In every book I can find, the Markov property for ito diffusions, $E[f(X_{t+h})\mid F_s] = E^{X_t}f(X_h)$ is stated for $\textbf{bounded}$ Borel functions.   However, I have the following statement from a set of notes: $$g(t,S_t) = E[(S_T-K)^{+}\mid F_t]$$  Where $g$ is some Borel function.  How can this be justified? Our $f$ here is $(x-K)^+$ which is not bounded.  Thank you
39003
Define the exponential Brownian martingale as $N_t = \exp\left\{a W_t - \frac12 a^2 t \right\}$ which is a martingale with respect to the natural filtration of $W$ which stands for a standard Brownian motion.  I now want to prove that $N_t \rightarrow 0$ a.s. for $t \rightarrow \infty$. As $N_t$ is a martingale it is by definition in $L^1$ hence Doob's martingale convergence theorem tells us that there exists a $N_\infty \in L^1$ such that $N_t \rightarrow N_\infty$ a.s., but how could I prove that $N_\infty \equiv 0$? Any help is greatly appreciated!
5426
Let $B_t$ be a 1-dimensional Brownian motion. I am following  by Bernt Øksendal. On the page 32 (it is displayed in the link I've put) there is a proof of existence of continuous version of the Ito integral. There is stated that for a function $$ \phi_n(t,\omega) = \sum\limits_j e_j(\omega)\cdot 1_{[t_j,t_{j+1})}(t) $$ its integral $$ I_n(t,\omega) = \int\limits_0^t\phi_n(s,\omega)dB_s(\omega) $$ is continuous in $t$.  From what I seen, I think that $$ I_n(t,\omega) = \sum\limits_{t_j\leq t}e_j(\omega)(B_{t_{j+1}} - B_{t_j}) $$ which does have jumps. So I wonder, how can it be continuous in $t$.
103016
can anyone help me with this: We are considering a symmetric random walk that ends if level 3 is reached or level -1 is reached. Start=0  What is the expected number of walks? So I am looking for: $E[{\tau}]$ with $\tau$=the stopping time.
26581
I have some discrete times of events and I would like to do a test to see if they are likely to have come from a homogeneous Poisson process.   From this pdf, I see:       REMARK 6.3 ( TESTING POISSON ) The above theorem may also be used to test the hypothesis that a given counting process is a Poisson process. This may be done by observing the process for a fixed time $t$. If in this time period we observed n occurrences and if the process is Poisson, then the unordered occurrence times would be independently and uniformly distributed on $(0, t]$. Hence, we may test if the process is Poisson by testing the hypothesis that the n occurrence times come from a uniform $(0, t]$ population. This may be done by standard statistical procedures such as the Kolmogorov-Smirov test.     However I don't quite understand what to do in practice.  Say my times are   [1, 7, 18, 22, 41, 43, 66, 73, 86, 92]   and the time interval I chose was from $1$ to $100$.  How exactly do I do the Kolmogorov-Smirov test in this example?
32897
I have two questions.   1) If $X_1,X_2,X_3,...,X_n$ constitute a random sample of size $n$ from an exponential distribution, show that $\bar X$ is a consistent estimator of the parameter $\lambda$.    This is my attempt: Mean of exponential distribution is $\lambda^{-1}$. Thus by central Limit theorem $E(\bar X)=\lambda^{-1}$. Using mean square consistency, if  $\lim_{n\to \infty} E[(\bar X -\lambda )^2]=0$ then $\bar X $ is consistent for $\lambda$. Since $$ E[(\bar X -\lambda )^2]= E[(\bar X)^2]+E(\lambda)^2-2\lambda E(\bar X) , $$  I get $$ \lim_{n\to \infty} E[(\bar X -\lambda )^2] =    \lim_{n\to \infty} {2 \over \lambda ^2 }+\lambda^2-2 . $$ Clearly this is not equal to $0$. How can I show this is consistent? I tried using Chebechev inequality as well. But even using that I couldn't.    2) Is $X_{(n)}$ a consistent estimator for $\lambda$? If $P[|X_{(n)}-\lambda|&lt;\epsilon]$ approaches $1$ as $n$ goes to infinity for all $\epsilon &gt; 0$ then $X_n$ is a consistent estimator for $\lambda$.    The probability density function of $X_{(n)}$ is $f_n(x) = n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}$,  for $X&gt;0$.  Thus \begin{align*} P[|X_{(n)}-\lambda|&lt;\epsilon]  &amp;= P[\lambda -\epsilon&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=P[0&lt;X_{n}&lt;\lambda+\epsilon] \\  &amp;=\int_0^{\lambda+\epsilon} n \lambda e^{-\lambda x}(1-e^{-\lambda x})^{n-1}\\  &amp;= (1-e^{- \lambda X_{(n)} }), \end{align*} where $X{(n)}$ takes values in $[0 , \lambda+ \epsilon]$.  Thus, can I say $\lim_{n\to \infty}[1-e^{-\lambda(\lambda+\epsilon)}]^n=1$?
108192
Let ${W_t}$ be 1 dimension Brownian motion and $X_t:=\exp(t/2)\cos W_t$ $t\in[0,T]$.  Show that $X_t$ is martingale.  I understood $df(t,W_t)=-\exp(t/2)\sin xdW_t$ , but I don't know why it become $X_t=1-\int_0^t \sin X_sdW_s$.  $f(0,X_0)=e^0 \sin X_0=1$? $X_0=?$  Please tell me.
15537
Let $B_t, t\ge0$ be a standard Brownian motion that starts at 0. Prove that, for any $\epsilon &gt; 0$,    $$P(\#\{t\in(0,\epsilon]\mid B_t = 0\}=\infty)=1.$$   This is a well-known fact, but surprisingly I cannot find a single proof on this site. The closest I get is , which proves the zero set of a standard Brownian motion restricted to $[0,1]$ is homeomorphic to the Cantor set. However, this proof seems to have utilised unfamiliar properties such as the ternary-expansion construction of the zero set.  Does there exist an elementary proof that only uses the familiar properties of the Brownian motion (normality, independent increments etc.)?
67571
1. On the internet, it is suggested that  $$ X_t=\left(\int_0^t X_s \;ds\right)\;dW_{t} $$ is a martingale, but not a Markov process. I understand that the process  $$ I_t(C)=\int_0^t C_s \; dW_s$$ is a martingale, but I don't see $X_t$ as an Ito's integral. In order to show it is a martingale but not a Markov process, I want to show that  $$ E(X_t\mid\mathcal{F}_{t-1})=X_{t-1}\neq E(X_t\mid\sigma(X_{t-1})). $$ I don't know how to take the expecation in this case.  Other counterexamples available online might be wrong. (Or, I could be wrong.)    2. Invalid counterexample: Let $e(t)$ be a random variable which is not independent of $y(0)$ and  $$ y(t)=y(t-1)+e(t)-E{[}e(t)|F(t-1){]}. $$  Then y(t) is a martingale but it is not a Markov process.  I will show that it is both a martingale and a Markov process. Define $X_{t}$ as  $$ X_{t}=X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right) $$  Then, $\{X_{t}\}$ is a martingale because $$ E(X_{t}|\mathcal{F}_{t-1})  =  E\left(X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\mathcal{F}_{t-1}\right)   =  E(X_{t-1}|\mathcal{F}_{t-1})+E\left(e_{t}|\mathcal{F}_{t-1}\right)-E\left(e_{t}|\mathcal{F}_{t-1}\right)  = E(X_{t-1}|\mathcal{F}_{t-1})   =  X_{t-1} $$  However, it is also a Markov process. $$ E(X_{t}|\sigma(X_{t-1})) =  E\left(X_{t-1}+e_{t}-E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right)   =  E(X_{t-1}|\sigma(X_{t-1}))+E\left(e_{t}|\sigma(X_{t-1})\right)-E\left(E\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right) $$  Since $\sigma(X_{t-1})\subset\mathcal{F}_{t-1}$, by the towering property, $$ E\left(\left(e_{t}|\mathcal{F}_{t-1}\right)\bigg|\sigma(X_{t-1})\right)=E\left(e_{t}|\sigma(X_{t-1})\right). $$  Thus $$E(X_{t}|\sigma(X_{t-1}))  =  E(X_{t-1}|\sigma(X_{t-1}))+E\left(e_{t}|\sigma(X_{t-1})\right)-E\left(e_{t}|\sigma(X_{t-1})\right)   =  E(X_{t-1}|\sigma(X_{t-1}) $$  Since $X_{t-1}$ is $\sigma(X_{t-1})$-measurable, $E(X_{t-1}|\sigma(X_{t-1}))=X_{t-1}$. Thus $$ E(X_{t}|\sigma(X_{t-1}))=X_{t-1}=E(X_{t}|\mathcal{F}_{t-1}) $$ Hence $\{X_{t}\}$ is also a Markov process.    3. Invalid counterexample from the textbook answer key. Alison Etheridge's textbook gives an example:  $$ S_{n+1}=S_{n}+\xi_{n+1} $$  where $\xi_{n}\in\{-1,1\}$ and $\{\xi_{n}\}_{n\geq0}$ are iid. Then, $$ P\left(S_{n+1}=k+1\bigg|S_{n}=k\right)=p \text{and }P\left(S_{n+1}=k-1\bigg|S_{n}=k\right)=1-p. $$ The author claims that $\left\{ S_{n}\right\} $ is a Markov process.  Then in the homework answer key, someone gives $\left\{ Z_{n}\right\}$ as a counterexample of a martingale not being a Markov process: $ \displaystyle{Z_{n}=\sum_{i=1}^{n}\xi_{i}} $  where $P(\xi_{i}=1)=\frac{1}{2}=P(\xi_{i}=-1)$.  Obviously, $Z_{n+1}=Z_{n}+\xi_{n+1}.$ So what's the difference between this counterexample and the example? Moreover, given $P(\xi_{i}=1)=\frac{1}{2}=P(\xi_{i}=-1)$, I think is both $\left\{ Z_{n}\right\}$ is both a martingale and a Markov process.
73772
Someone please help me with this:  I have that $X_t$ is the Ito diffusion with genertator $A(f)(x)=\alpha xf'(x)+f''(x).$   Then, if $X_0=x \in \mathbb{R}^+$, how do I find $X_t$?
42784
Let $X_1$ and $X_2$ be independent and identically distributed exponential random variables with rate $\lambda$. Let $S_2 = X_1 + X_2$.  Q: Show that $S_2$ has PDF $f_{S_2}(x) = \lambda^2 x \text{e}^{-\lambda x},\, x\ge 0$.  Note that if events occurred according to a Poisson Process (PP) with rate $\lambda$, $S_2$ would represent the time of the 2nd event.   Alternate approaches are appreciated. The approaches provided are commonly used when learning queueing theory &amp; stochastic processes.     Recall the Exponential distribution is a special case of the Gamma distribution (with shape parameter $1$). I've learned there is a more general version of this  that can be applied.
82222
Addressing this problem:  Let $X_1, ..., X_n$ be iid exp($\theta$) with pdf,  $f(x)=\frac{1}{\theta}e^{-\frac{x}{\theta}}$. What is the distribution of the sample variance?  This is to be done by proof, not by simulation/sampling. Everything online says simulation/sampling, so is there no formal mathematics for determining this distribution?
93005
Letting $dt$ be an infinitesimal interval, what is the argument to that $$f(t | H_{t_n})\;dt = P (t \in [t,t+dt] | H_{t_n}),$$ where $H_{t_n}$ denotes the history of the previous points before $t$? I know that the probability to be in a certain interval is by integrating the density function over that interval. But why is it in this case equal to the density function times the length of the interval? I've gotten an explanation that it's because we can assume that the density function is constant, as we are considering an infinitesimal interval. However, I can see that the equality then is true if the pdf is constant, but I don't understand why we can assume that the it is constant when considering an infinitesimal interval.
41108
I'm trying to prove this conclusion but have some problems with one of the steps.  Assume $X_1,\ldots,X_n,\ldots$ is a sequence of Gaussian random variables, converging almost surely to $X$, prove that $X$ is Gaussian.  We use characteristics function here. Since $|\phi_{X_n}(t)|\leq 1$, by dominated convergent theorem, we have for any $t$  $$ \lim_{n\rightarrow\infty}e^{it\mu_n-t^2\sigma_n^2/2}=\lim_{n\rightarrow \infty}\phi_{X_n}(t) = \lim_{n\rightarrow \infty}\mathbb{E}\left[e^{itX_n}\right] = \mathbb{E}\left[e^{itX}\right] = \phi_X(t) $$  this is the step that I cannot figure out: $e^{it\mu_n-t^2\sigma_n^2/2}$ converges for any $t$ if and only if $\mu_n$ and $\sigma_n$ converges.   Let $\mu=\lim_n \mu_n$, and $\sigma=\lim_n\sigma_n$, then $\phi_X(t)=e^{it\mu-t^2\sigma^2/2}$, which proves that $X$ is a Gaussian random variable.  Why can we get that $\mu_n$ and $\sigma_n$ converge? This looks intuitive for me, but I cannot make a rigorous prove.
69139
Let $B_t$ be the standard Brownian motion process, $a &gt; 0$, and let $H_a = \inf \{ t : B_t &gt; a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is $$\mathbb{E}[\exp(-\lambda H_a)] = \exp (-\sqrt{2\lambda} H_a)$$ by considering the martingale $$M_t = \exp \left(\theta B_t -\frac{1}{2}\theta^2 t\right)$$  There's an obvious argument to follow here: assuming the optional stopping theorem applies, we have $$1 = \mathbb{E}[M_{H_a}] = \mathbb{E} \left[ \exp \left(\theta a - \frac{1}{2}\theta^2 H_a\right) \right] = \exp(\sqrt{2\lambda} a) \mathbb{E} \left[ \exp(-\lambda H_a) \right]$$ where $\theta = \sqrt{2\lambda}$. This is exactly what we wished to show. However, as far as I can tell, the hypotheses of the optional stopping theorem are not satisfied here. Here is the statement I have:     If $(X_n)$ is a martingale and $T$ is an a.s. bounded stopping time, then $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.   I think not all is lost yet. $M_t &gt; 0$ for all $t$, so the martingale convergence theorem applies, and $M_t \to M_\infty$ a.s. for some integrable random variable $M_\infty$. For each $t$, $H_a \wedge t = \min \{ H_a, t \}$ is a bounded stopping time, so certainly $\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_0]$. But, $$\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] + \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}]$$ and clearly what one wants to do is to take $t \to \infty$ on both sides. But here's where I get stuck: I'm sure I need a convergence theorem here in order to conclude that the equation remains valid in the limit.   Now, $0 &lt; M_{H_a} = \exp \left(\theta a - \frac{1}{2} \theta^2 H_a \right) \le \exp(\theta a)$, so the dominated convergence theorem applies, and so $$\lim_{t \to \infty} \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a &lt; \infty\}}]$$ and I believe Fatou's lemma gives me that $$\liminf_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}] \ge \mathbb{E}[M_{\infty} \mathbf{1}_{\{H_a = \infty\}}]$$ but I think what I need is the equality $$\lim_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a &gt; t\}}] = \mathbb{E}[M_\infty \mathbf{1}_{\{H_a = \infty\}}]$$ and as far as I can tell neither the monotone convergence theorem nor the dominated convergence theorem applies here. Is there anything I can do to rescue this line of thought?