New answers tagged stochastic-calculus
1
Your equation looks ok. If interest rates are deterministic then forwards (being the same as futures) are driftless under the risk neutral measure. Otherwise, Forwards are driftless (i.e. martingales) under the corresponding forward measure while futures are martingales under the risk neutral measure.
1
Let $$Z_t=Y_t\int_0^t\frac{a}{Y_s}ds$$
Then $Z_0=0$.
We differentiate $Z_t$ and obtain
$$dZ_t=\int_0^t\frac{a}{Y_s}dsdY_t+Y_t\frac{a}{Y_t}dt=\int_0^t\frac{a}{Y_s}ds(rY_tdt+\sigma Y_td\tilde{W_t})+adt$$
$$=rY_t\int_0^t\frac{a}{Y_s}dsdt+\sigma Y_t\int_0^t\frac{a}{Y_s}dsd\tilde{W_t}+adt$$
Then
$$dZ_t=rZ_tdt+\sigma Z_td\tilde{W_t}+adt$$
We have
$$d(e^{-rt}Z_t)=...
0
$Value(t)$ is the value at time $t$ of receiving \$1 at time $T$. Thus, indeed
$$Value(t)=\exp\left(-\int_t^T r_u\mathrm{d}u\right).$$
This expression is also known as (stochastic) discount factor. Let $Wealth(t)=\exp\left(\int_0^t r_u\mathrm{d}u\right)$ be the value at time $t$ of investing \$1 at time $0$. Your \$1 grows at the stochastic rate $r_t$. Then,...
2
Let's take a standard Brownian motion $(B_t)$ and let's try to compute $\int_0^t B_s\mathrm{d}B_s$ in the Riemann-Stieltjes sense.
Let $0=t_0<t_1<...<t_n=t$ be a partition and let $y_i=t_{i-1}$ or $y=t_i$ for $i=1,...,n$ be two intermediate partitions. Thus,
\begin{align*}
S^1_n(t) &= \sum_{i=1}^n B_{t_{i-1}}(B_{t_i}-B_{t_{i-1}}), \\
S^2_n(t) &...
1
Yes, different banks using different models will get different Greeks. Some of them will be right and some of them will be wrong. What do we mean by right and wrong? ‘Right’ means that when the market moves, your Greeks closely predict how the market values of the options are moving. There are multiple examples in all asset classes (rates, equities, fx) ...
1
For any price process $Z$, given that $N$ and $M$ are numeraire processes,
\begin{align*}
E_N\left(\frac{Z_T}{N_T} \right) = E_M\left(\frac{Z_T}{M_T}\frac{M_0}{N_0} \right),
\end{align*}
as they both equal to $Z_0/N_0$. Note also that
\begin{align*}
E_N\left(\frac{Z_T}{N_T} \right) = E_M\left(\frac{dQ_N}{dQ_M}\frac{Z_T}{N_T} \right).
\end{align*}
Then
\...
2
I assume you are interested in the bond price. Let $B_t=B(t,r_t)=\mathbb{E}^\mathbb{Q}[\exp\left(-\int_t^T r_u\mathrm{d}u\right)\mid\mathcal{F}_t]$ be the time $t$ price of a default-free zero-coupon bond maturing at $T$.
That is, by the way, very different to $\exp\left(-\int_0^t r(u)\mathrm{d}u\right)$ which relates to a money-market account (bank account,...
4
Any nonnegative random variable $Z$ with expectation 1 is a Radon-Nikodym derivative:
$$
\mathbb{E}^{\mathbb{P}} \left(Z\right) = \mathbb{E}^{\mathbb{P}} \left(\frac{\mathrm{d}\mathbb{Q}}{\mathrm{d}\mathbb{P}}\right) = \mathbb{E}^{\mathbb{Q}} \left(1\right) = \int{\mathrm{d}\mathbb{Q}} = 1
$$
$$
\mathbb{Q} \left(A\right) = \mathbb{E}^\mathbb{P} \left(Z 1_A\...
0
Direct implication is not true. Function
$$f(x) = \left\{\matrix{x^2\sin\left(\frac{1}{x^4}\right) & x \not= 0\\0 & x = 0}\right.$$
is continuous, but its quadratic variation over interval $[0,1]$ is non-zero. You can find a proof here.
(Bounded variation and differentiability of functions of type $y=x^a \sin (1/x^b)$ can be found here and here.)
3
The statement is not quite correct. Brownian motion has continuous sample paths but not zero quadratic variation. In fact, $[B_t]=t$, which is finite yet unbounded. The sample paths of Brownian motion have infinite variation which is why we need the Itô integral in the first place and can't simply use the Riemann-Stieltjes integral.
You need $f$ to be of ...
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