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## Hot answers tagged stochastic-calculus

3

The original renewal process has independent and identically disributed (i.i.d.) inter-renewal times $\{T_1, T_2, T_3, ...\}$. Thus, starting from time 0, renewals occur at times $\{T_1, T_1+T_2, T_1+T_2+T_3, ...\}$. Now fix a probability $p >0$. Independently place each renewal time to $P_1$ with prob $p$. So we get new inter-renewal times $\{Z_1, ... 3 This is a partial answer, just too long for a comment. There is a nice lemma from Dellacherie and Meyer: Lemma If$\{A_s,s\in [0,t]\}$is a continuous non-decreasing adapted process such that$E[A_t - A_s|\mathcal F_s]\le K$for all$s\in[0,t]$, then for any$\lambda < 1/K$, $$E[e^{\lambda A_t}]< (1-\lambda K)^{-1}.$$ Now take$A_s = ...

2

Hint: Your argument shows that $\Bbb P(\tau>km)\le(1-\epsilon)\Bbb P(\tau>(k-1)m)$ for $k=1,2,\ldots$. (Take your $n$ equal to $k-1$.) Proceed recursively.

2

There are lots of random variables that are positive with probability $1/2$ and negative with probability $1/2$, but their expected value is not $0$. But the point about $\sin$ being odd is a good one. What it means is that the distribution of your random variable is symmetric about $0$, and that does imply that the expected value (if it exists) is $0$.

1

As you correctly write, the integral $\int_0^t b(X_s) dX_s$ is well defined (as an extended Itô integral) provided that $\int_0^t b(X_s)^2 ds<\infty$ almost surely. So the question is: When is the integral $\int_0^t f(X_s) ds$ finite almost surely? The answer to the latter question is well known: this is the case if$^*$ $f\in ... 1 You don't have the right argument, the real reason why you can get the third line is because first$B$and$Z$are orthogonal so that $$\mathbb{E}^{\mathbb{Q}^{t} }\bigg[\bigg(\int_0^t\sigma_r^t(s)dB_s^t \bigg)\cdot\bigg(\int_0^t\mathbb{E}^{\mathbb{Q}^{t}}_s[D_s^{Z^t}\mathbb{1}_{S_t > K}]dZ_s^t \bigg)\bigg] =0$$ and because : ... 1 Use the integrating-factor approach, where$\mu_t = e^{-a t}$. Rewrite the original SDE as: $$d\left( \mu_t Y_t \right) = \mu_t \, b \, dX_t \, .$$$Y_t\$ can be obtained as $$Y_t = \frac{b}{\mu_t} \int_0^t \mu_s \, dX_s = b\, e^{-at} \int_0^t e^{a s} dX_s \, .$$

1

I think you have nearly answered your own question. The finite nature of the inequality in (ii) follows from the finite property of the expectation of the supremum, i.e. $$\int\limits_{ \{ 0\leq T \leq t \} } |X_T| \ dP \leq E\left[ \sup_{ 0\leq u \leq t } |X_u|\right] < \infty,$$ and hence, $$E\left[ |X_{T\wedge t}| \right] \leq E\left[ \sup_{ 0\leq ... 1 By definition a random variable X is just a measurable function, hence it does not need to be integrable. Anyway, if you assume that X is integrable, then (by DCT)$$ \lim_n \int f_n \mathrm{d}P = \int \lim_n f_n \mathrm{d} P. $$whenever f_n \le |X|. Then, it is enough to choose f_n=X \cdot {1}_{[n,\infty)}(|X|). 1 Note that$$Y_t = \alpha A e^{t^2/2} + e^{t^2/2} B_t$$has to satisfy the initial condition Y_0 = \alpha. Thus,$$Y_0 = \alpha A \stackrel{!}{=} \alpha,$$i.e. A=1. Consequently,$$Y_t = e^{t^2/2} (\alpha+B_t).$$If you want to check whether this is indeed a solution to the given SDE, then just apply Itô's formula with$$f(t,x) := e^{t^2/2} ...

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