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Define $X(t) = \int_0^t \sigma(s) dW(s)$ where $W$ is a Wiener process aand $\sigma(s)$ some deterministic process. Informally this is the same as $dX(s) = \sigma(s) dW(s)$.

I wish to show that $X(t)$ is normally distributed with $0$ mean and variance $\int_0^t \sigma^2(s)ds$.

By letting $Z(t) = e^{u X(t)}$, we obtain from Ito's formula that $$dZ(t) = \frac{1}{2} \sigma^2(t) u^2 Z(t)dt + \sigma(t)uZ(t) dW(t). $$ In integral form, this is $$Z(t) = 1 + \int_0^t \frac{1}{2} \sigma^2(s) u^2 Z(s) ds + \int_0^t \sigma(s) uZ(s) dW(s). $$ Then we take expectations and after that, we differentiate in order to get a solvable DE.

First problem: Can I just move an expectation inside the first integral above?

Second problem: Is the expectation of the second integral 0? I know this to be true if the integrand is an adapted process which satisfies a particular integrability condition, but I am unsure if all that is satisfied here?

Third problem: Assuming the answers to above are yes, yes, we then want to take the derivative on both sides. On the right, the constant disappears but the integral remains. How do I take its derivative? Informally, I would presume it equals $\frac{1}{2} \sigma^2(t) u^2Z(t)$. Is this true?

Fourth problem: If the answer to above is yes, how do I solve the DE? I am unsure what to do with $\sigma$ when it depends on $t$.

ImeanH
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  • You have stated that informally this is the process $dX(s) = \sigma(s)dWs$, of course this is no more informal then any other presentation of an SDE, I think maybe you have introduced some confusion with this new variable $Z$, but all the properties you want come from the Ito integral itself. For example you can find the variance by applying the Ito isometry result if you have seen this? – Nadiels Dec 02 '16 at 22:15
  • Have a look at the answer to this question http://math.stackexchange.com/questions/308120/finding-the-distribution-of-an-ito-integral-int-0t-sb-s-mathrmds?rq=1 - which is the special case when you function $\sigma(s) = s$, have a look at the accepted answer and see how you can adapt those results to the more general, but still deterministic, case. – Nadiels Dec 02 '16 at 23:38

2 Answers2

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I do not know why you are studying $Z_t$, it sounds redundant... by construction you integral is normal. You know exactly what is $Z_t$.

It is an Ito integral with a deterministic integrand on the Brownian side, therefore it is normal. The Ito isometry https://en.wikipedia.org/wiki/It%C3%B4_isometry can help you to get the variance.

First problem : yes, if you assume intregrability. Second problem : it exists if $E(\int_0^t {(\sigma(s)uZ(s))}^2 ds) <\infty$

Third and four :I am not sure what you mean, but refer to the first lines I wrote

Canardini
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    This is not an answer to the questions being asked. It should be a comment. – Jaood Dec 02 '16 at 20:55
  • Read the question : show that X is normal , not clear ? He is trying to prove that X is normal using Ito's lemma on a more complicated function. – Canardini Dec 02 '16 at 21:01
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    Yeah in defense of @Canardini he has addressed the meat of the problem - normality of the Ito integral, the problems in bold really stem from the unnecessary complexity introduced by the transformation to this new processes $Z_t$, which looks like is just an attempt to transform the given $X(t)$ to a process of the form $dZ(t) = fdt + gdW_t$, because this is how these things get presented, when really the object of interest is the Ito integral $\int_0^t \sigma(s) dW_s$. – Nadiels Dec 02 '16 at 21:13
  • @Nadiels: The meat of the problem is what OP spent 90 % of his question on, which is the introduction of Z and its stochastic differential. Canardini then proceeded to ignore all of that, and (poorly) gave a different answer, which clearly uses theory the OP has yet to study - if the OP knew of the knowledge presumed in Canardini's answer, the OP would not have asked this question in this shape to begin with. – Jaood Dec 02 '16 at 21:49
  • @Jaood Ok I'm new to all this so I'm not going to get overly involved here and I don't know where the line is drawn as far as responders answering the question they believe was intended which I still maintain is the normality of the Ito integral and not the questions about adaptedness and measurability and so on of this $Z_t$ process. With that in mind I would agree that the answer was terse and not wonderfully set out - but he still points out that the normality follows from propertys of the Ito integral and surely the OP has begun the study of that theory? – Nadiels Dec 02 '16 at 22:10
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It is correct that we obtain $$Z(t) = 1 + \int_0^t \frac{1}{2}\sigma(s)^2u^2 Z(s) ds + \int_0^t \sigma(s)u Z(s) dW(s), $$ $$ \Longrightarrow E(Z(t)) \equiv g(t) = 1 + \int_0^t \frac{1}{2}\sigma(s)^2 u^2 g(t) + \int_0^t \sigma(s)ug(t)dW(s),$$ $$\Longrightarrow g'(t) = \sigma(t)^2 \frac{1}{2}u^2 g(t) \Longrightarrow g(t) = e^{\int_0^t \frac{u^2}{2}\sigma(s)^2 ds}$$

Therefore, $E( e^{u X(t)}) = e^{\int_0^t \frac{u^2}{2}\sigma(s)^2 ds}.$ Now recall that a normally distributed variable $N$ with mean 0 and variance $\sigma^2$ has $E( e^{u N(t)}) = e^{ \frac{1}{2}u^2 \sigma^2 }.$ By comparison, we see that $X(t)$ is normally distributed with mean 0 and variance $\int_0^t \sigma(s)^2 ds$.

We used in the above calculations integrability and that the Ito-integral disappears once we take expectations. See your problem formulation in order to determine whether these assumptions are satisfied. They depend on the properties of $\sigma(s)$.

Jaood
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    If you are really arguing that this is the question the OP intended to ask then fair enough, I still believe he meant to ask the question in the heading and his second line... you argue that since the majority of his text was about the process $Z_t$ then this is the important feature, and I'm not sure that is the case, but there we go I'm sure the OP will clear that up. – Nadiels Dec 02 '16 at 22:23