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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 ...
Kevin's user avatar
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10 votes
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List: Behavioural characteristics of key Ito processes used in finance

I will provide some references such that you can see where the different processes are used. These papers typically motivate their models and show which effect the single paramaters have and what ...
10 votes
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Deriving the solution for European call option in the Heston Model

Itô's Lemma The standard version of Itô's Lemma applies to a single Itô process $\text{d}X_t=\mu(t,X_t)\mathrm{d}t+\sigma(t,X_t)\mathrm dW_t$. Then, $$\mathrm{d}f(t,X_t) = \left(f_t+\mu(t,X_t)f_x + \...
Kevin's user avatar
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9 votes

More questions about integral of Brownian Motion w.r.t time

It is indeed Riemann integrable, so you don't need stochastic integration. For a given path, you can interpret the integral in the Riemann sense. For a given t, the paths are random, so it is a random ...
Magic is in the chain's user avatar
9 votes
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More questions about integral of Brownian Motion w.r.t time

As usual with those kind of integrals, another way to reach the result is to: Express $W_s$ in integral form as $\int_0^s dW_u$ Use Fubini theorem to change the integration bounds of the resulting ...
Quantuple's user avatar
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9 votes
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Gamma PnL from Itô's Lemma derivation

$$ \frac{1}{2} \frac{\partial^2 f}{\partial S^2} dS^2 \approx \frac{1}{2} \sigma^2 S^2\frac{\partial^2 f}{\partial S^2} dt$$ (for small $dt$, ignoring $(dt)^2$ terms ) $\sigma$ is embedded in $dS = \...
ir7's user avatar
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8 votes
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2 Ito processes - $d(X_{t} + X^{'}_{t})^2 = (Y_t + Y^{'}_{t})^2 dt$ why it is true?

$X_t$ being a stochastic process, one cannot use ordinary calculus to express the differential of a (sufficiently well-behaved) function $f$ of $t$ and $X_t$. Instead one should turn to Itô's lemma, ...
Quantuple's user avatar
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8 votes

How can I learn stochastic process & stochastic calculus in two weeks?

This is impossible unless you are very intelligent with good memory-retention skills and already mathemathically proficient in the field of analysis and statistics (and no, a single course in basic ...
saei's user avatar
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7 votes
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Baxter & Rennie HJM: differentiating Ito integral

Let $$Z_t = \exp(-X_t)$$ with $$X_t = \sigma(T-t)W_t+\sigma\int_0^tW_sds+\int_0^Tf(0,u)du+\int_0^t\int_s^T\alpha(s,u)du ds $$ and $W_t$ a standard Brownian motion, along with the usual assumptions. ...
Quantuple's user avatar
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7 votes
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Why is $S(t) = e^{\alpha + \beta t + \sigma W(t)}$ used as a model for prices?

We don't model the prices, we model the returns. The stock prices aren't explicitly modelled as log-normal, but rather this is a consequence of the actual model used to describe the returns. The core ...
oliversm's user avatar
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7 votes

Can I write Ito's Lemma as a taylor expension?

Ito Lemma (as 'Taylor expansion'): For $X$ an Ito process and $f = f(t, x) ∈ C^{1,2}(\mathbb{R}^2)$ a deterministic function, the stochastic process $$Y_t = f(t,X_t)$$ is an Ito process and we have $$...
ir7's user avatar
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7 votes
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conditional expectation of stochastic integral

What a great question! I've had a go at it below, I'd say I'm about 75% sure of the result I've got to but I'd love feedback from others. I'm going to use the definition of the Ito integral, \begin{...
StackG's user avatar
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7 votes

Ito calculus is Gaussian (using method of characteristic function)

Hints: First show (using Ito Lemma) that $$ \exp(iuX_t) = 1 + iu \int_0^t\exp(iuX_s) h(s) d W_s -2^{-1}u^2 \int_0^t\exp(iuX_s) h(s)^2ds$$ Then show (by taking expectations): $$ E[\exp(iuX_t)] = 1 - 2^...
ir7's user avatar
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7 votes
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Ito multiplication

If $M$ and $N$ are independent (your references appear to make this assumption), then $M+N$ is also a Poisson process. So, using the polarization identity: $$ dMdN = 2^{-1}\left[(d(M+N))^2 - (dM)^2 - (...
ir7's user avatar
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6 votes
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Quantile normal and lognormal

Quantiles are preserved under monotonic transformations, hence the quantile for $Y$ is simply the exponential of the quantile of $X$, no need for corrections whatsoever (see here for instance). Put ...
Quantuple's user avatar
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6 votes
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Stochastic differential equation of a Brownian Motion

In stochastic calculus, only stochastic integrals are defined. The differential form is just a notation. That is, $$dF=g(t)dW_t$$ is just another expression for the integral $$F=\int_0^t g(s) dW_s.$$ ...
Gordon's user avatar
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6 votes

How is the Wiener integral $\int{WdW}$ calculated?

You are "deriving" with respect to $t$ (the time index in your stochastic process). $f(t,x) = x^2$ so $f(t,W_t) = W_t^2$. And Ito's lemma tells you $W_b^2 - W_a^2 = \int_{t=a}^b d(W_t^2) = \int_{t=...
AFK's user avatar
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6 votes
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exercise on multivariate Ito's lemma + jumps (Poisson)

Answer Assuming the Poisson process $N_t$ is independent from the Brownian motions $(W_{1,t},W_{2,t})$, you'll have \begin{align} df(X_{1,t},X_{2,t}) &= \frac{\partial f}{\partial X_{1,t}} dX_{1,...
Quantuple's user avatar
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6 votes

Basic question on Ito integrals

To verify @AntoineConze's suggestion, the variance should be: $$\int_0^4 (2_{[0,1]}(t)+3_{(1,3]}(t)-5_{(3,4]}(t))^2\,dt.$$ Since the supporting domains are disjoint, the product of any two of the ...
Bjørn Kjos-Hanssen's user avatar
6 votes
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Application of Ito's lemma

Consider OP's general formula $f(g(t),X_t)$. In case of ambiguity, let us claim that $f=f(t,x)$ is defined with variables $t$ and $x$, $g=g(s)$ is defined with the variable $s$, and $h=h(u,x)=f(g(u),...
hypernova's user avatar
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6 votes
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Ito`s Lemma problem

write down Ito's lemma for the function X: $$dX=\frac{\partial X}{\partial Y}dY+\frac{1}{2}\frac{\partial^2 X}{\partial Y^2}(dY)^2+\frac{\partial X}{\partial c}dc+\frac{1}{2}\frac{\partial^2 X}{\...
ZRH's user avatar
  • 1,671
6 votes

Variance of a time integral with respect to a Brownian Motion function

Using Fubini's argument, assuming that $f$ is deterministic $$E(I_t^2) = E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)=\int_0^t\int_0^t{f(s)f(u)min(s,u)duds}$$ If $f$ is continuous(even ...
Canardini's user avatar
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6 votes

Variance of a time integral with respect to a Brownian Motion function

As @Canardini pointed out, \begin{align*} E\big(I_t^2\big) &= E\left(\int_0^t f(s) W_s ds\int_0^t f(u) W_u du\right)\\ &= \int_0^t\!\int_0^t f(s)f(u)\min(s,u)dsdu\\ &= \int_0^t\left(\int_0^...
Gordon's user avatar
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6 votes
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Pricing call option using risk-neutral martingale approach with squared stock price boundary?

You do not really need the dynamics of $S_t^2$. You can simply apply your standard technique from risk-neutral pricing. The time zero price of a European-style contract with payoff $X$ is given by $$...
Kevin's user avatar
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6 votes

Application of Ito's Lemma in expected utility theory

The risky and riskless assets follow processes, $$\frac{dS_t}{S_t}= \mu \, dt + \sigma \, dB_t, \,\,\, \frac{dM_t}{M_t}= r \, dt$$ If the proportion invested in the risky asset at time $t$ is $p_t$, ...
RRL's user avatar
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6 votes

Itos Lemma Derivation notation

The theory behind the actual reasoning is a bit complicated than the coverage in Hull's, but staying within the simple reasoning, the difference comes down to the following: The Brownian increments ...
Magic is in the chain's user avatar
6 votes

Derivative of Stochastic Integral

No. Itō’s formula helps you derive the dynamics of $f (S_\cdot )$ given the SDE followed by $S$. Here this is not the case. You simply have: $$ \mathrm{d} \left[\int{g(S_t)\mathrm{d}S_t}\right] = g(...
siou0107's user avatar
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6 votes
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Proving that a stochastic process is a martingale using Ito's Lemma

$$ d Y \left(t\right) := d \left[\int_0^t{a \left(s\right)\mathrm{d}W_s}\right] = a \left(t\right) dW_t $$ Note that since $Y$ is a driftless process, it is a local martingale, and because $a$ is ...
siou0107's user avatar
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