55 votes

Integral of Brownian motion w.r.t. time

This type of integral has appeared so many times and in so many places; for example, here, here and here. Basically, for each sample $\omega$, we can treat $\int_0^t W_s ds$ as a Riemann integral. ...
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24 votes
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Worked examples of applying Ito's lemma

These are all examples on Ito Formula in its general form (with quadratic variations):
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18 votes
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Why is Brownian motion useful in finance?

Brownian motion is simply the limit of a scaled (discrete-time) random walk and thus a natural candidate to use. It is very intuitive and arguably one of the simplest and best understood time-...
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  • 13.8k
16 votes
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Why is this stochastic integral a martingale?

In the integral $$\int_0^t S_u dW^{*}_u \, ,$$ $dW^{*}_u \equiv W^{*}_{u+du} - W^{*}_u$ is independent from the integrand $S_u$. So, $\mathbb{E}\left[ \int_0^t S_u dW^{*}_u\middle\vert \mathcal{F}...
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  • 1,118
15 votes

What is the trickiest thing to get right in Rates Quant recently (2019)?

Of course making money is always the key issue. That (not completely facetious) comment aside: On the practical side, in many firms IT is struggling with being clear, transparent, and intuitive in ...
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13 votes
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Girsanov Theorem for Quanto/Compo adjustment

Assume deterministic and constant interest rates. For an investor in the foreign economy i.e. a market participant that can only trade assets delivering a payout in the foreign currency, let us define ...
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13 votes
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Solve the following SDE: $\mathrm{d}X_t = a(b-X_t) \,\mathrm{d}t + c X_t \, \mathrm{d}W_t$

Let \begin{align*} Y_t = e^{(a+\frac{c^2}{2})t-cW_t}. \end{align*} Then \begin{align*} dY_t = Y_t\left[\big(a+c^2\big)dt -c dW_t \right]. \end{align*} Moreover, \begin{align*} d(X_tY_t) &= Y_t ...
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13 votes
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Clarification on Deriving Ito's Lemma

Just a few notes How to make sense of $\text dW_t$ is the entire point of stochastic calculus. It's far beyond the scope of any answer here. You should read some introductory lecture notes/books on ...
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12 votes

Geometric Brownian motion - Volatility Interpretation (in the drift term)

The convexity of the exponential function of the stochastic variable $W$ makes its expectation greater than the exponentiation of the expectation of $W$. This is an example of Jensen's inequality, $E[...
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  • 2,473
12 votes

Why the expected return rate of a stock has nothing to do with its option price?

Because you can hedge. Once you have delta hedged, the pay-off is symmetric about up and down moves so drift doesn't matter. Also the delta-hedged call and the delta hedged put have to have the same ...
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  • 6,743
12 votes

Integral of Brownian motion w.r.t. time

Just to add to the already nice answers, the result can also be obtained using the (stochastic) Fubini theorem. \begin{align} \int_0^t W_s ds &= \int_0^t \int_0^s dW_u\, ds \tag{$W_s=\int_0^s ...
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12 votes
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Can I always use quadratic variation to calculate variance?

Quadratic variation and variance are two different concepts. Let $X $ be an Ito process and $t\geq 0$. Variance of $X_t$ is a deterministic quantity where as quadratic variation at time $t $ that ...
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  • 2,362
11 votes
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Geometric Brownian motion - Volatility Interpretation (in the drift term)

I will try to answer this a bit differently. The rigorous answer: because Ito calculus tells us that we need the second order term. Look at $$ S_t = S_0\exp(\mu t + \sigma B_t). $$ Assume that $S_0$ ...
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11 votes

How to use the stock as a numeraire to price a derivative with payoff of the form $(S_T f(S_T))^+$?

Let $P$ be the risk-neutral measure. We define the measure $P_S$ such that \begin{align*} \frac{dP_S}{dP}\big|_t &=\frac{S_t}{e^{rt}S_0}\\ &=e^{-\frac{1}{2}\sigma^2 t+\sigma W_t}. \end{align*} ...
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11 votes

Why Ito calculus?

My understanding is because the Ito's integration definition keeps the martingale property. With Brownian motion $W(t, \omega)$ defined, to define stochastic integration in a Riemann–Stieltjes style:...
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11 votes
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Ho and lee derivation for short rates model

For any $s \geq t$, note that \begin{align*} r_s = r_t + \sigma\int_t^s dW_u + \int_t^s \theta_u du. \end{align*} Then, \begin{align*} \int_t^T r_s ds &= (T-t)r_t + \sigma\int_t^T\int_t^s dW_u ds +...
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11 votes
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How were these SDE derived?

From $(2)$, \begin{align*} \ln S_t &=\ln F_{t, t} \\ &= \ln F_{0, t}-\frac{1}{2}\int_0^t\sigma^2 e^{-2\lambda (t-s)}ds+\int_0^t \sigma e^{-\lambda(t-s)} dB_s\\ &=\ln F_{0, t}-\frac{\sigma^...
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11 votes
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What is an adapted process

Let $\{X_t\}$ be a stochastic process and $\mathcal{F}$ be a filtration. The intuitive idea is that for $\{X_t\}$ to be adapted, it can't reveal what's unknowable (according to the filtration). By ...
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  • 6,294
11 votes

Two papers - two different solutions of the Ornstein-Uhlenbeck process

Note that the Ito integral of a deterministic integrand $f: \mathbb{R}_+ \rightarrow \mathbb{R}$ is normally distributed \begin{equation} \int_0^t f(u) \mathrm{d}W_u \sim \mathcal{N} \left( 0, \...
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11 votes
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Expectation of $\int_0^t \frac{1}{1+W_s^2} \text dW_s$

By construction, the Itô integral, $I_t=\int_0^t X_s\text{d}W_s$, is a martingale if $\int_0^t \mathbb{E}[X_s^2]\text{d}s<\infty$. The martingale property, $\mathbb{E}_s[I_t]=I_s$ implies $\mathbb{...
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11 votes

Expectation of exponential of 3 correlated Brownian Motion

You need to rotate them so we can find some orthogonal axes. A simple way to think about this is by remembering that we can decompose the second of two brownian motions into a sum of the first ...
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11 votes
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Expectation of exponential of 3 correlated Brownian Motion

Besides @StackG's splendid answer, I would like to offer an answer that is based on the notion that the multivariate Brownian motion is of course multivariate normally distributed, and on its moment ...
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10 votes
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Girsanov Theorem and Quadratic Variation

Shreve's theorem also called "Girsanov II" indeed represents a special case of the general "Girsanov I" from Wiki above, with $$Y_t:=W_t,$$$$X_t:=-\int_0^t\Theta_udW_u$$ We can show: $$[Y,X]=-\int_0^...
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10 votes

Why is this stochastic integral a martingale?

I aim to give a careful mathematical treatment to this answer, whilst following the fantastic book "Basic Stochastic Processes" by Brzezniak and Zastawniak. The reason I am putting this answer on is ...
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10 votes
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Correlation coeffitiont between two stochastic processes

if you talk about correlation then: compute expectation: $$\mathbb{E}(W_t)=0\text{ and }\mathbb{E}(\int_0^tW_d ds)=0$$ variance: $$\text{Var}(W_t)=t\text{ and }\text{Var}(\int_0^tW_s ds)=\frac{t^3}{...
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  • 2,362
10 votes

Why is Brownian motion useful in finance?

Physical objects move according to simple smooth curves that can be represented by low order polynomials: a straight line, a parabola, an ellipse, etc. Financial market prices move in a completely ...
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  • 9,077
10 votes
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Periodic functions when determining No Arbitrage price

It is, of course, possible to price such a contract in a no-arbitrage market. Indeed, if $f$ is a sufficiently smooth function, then you can price all contracts paying $f(S_T)$. Note that your ...
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