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I have recently come across an attempt to differentiate the following function with respect to $t$ and with respect to $W_t$:

$$F(W_t, t):=\int_{h=0}^{h=t}W_hdW_h$$

Is it possible (i.e. is it "well defined") to differentiate $F$ with respect to $t$ and with respect to $W_t$?

My train of thought would be as follows: by applying Ito's lemma to $W_t^2$, we know that $F_t$ evaluates to $\frac{1}{2}(W_t^2-t)$. Differentiating with respect to $W_t$ would then simply yield $W_t$, whilst differentiating with respect to $t$ would then yield $-\frac{1}{2}$:

Question 1: is the above correct? (I have doubts because $\frac{\partial W_t}{\partial t}$ is undefined. But on the other hand, we are effectively carrying out $\frac{\partial F}{\partial t}$, so following the logic of partial derivatives applied to a function of multiple variables, we should be able to treat all variables "not being differentiated with respect to" as constants).

Question 2: In general, is there a "rule" (that would be an "Ito" equivalent of the Fundamental Theorem of Calculus) for differentiating an Ito integral with respect to its upper limit?

Question 3: What if I now define the function $F(W_1(t),W_2(t),t)$ as follows ($W_1$ and $W_2$ are two independent Brownian motions on the same probability space):

$$F(W_1(t),W_2(t),t):=\int_{h=0}^{h=t}W_1(h)dW_2(h)$$

Would $\frac{\partial F}{\partial t}$, $\frac{\partial F}{\partial W_1}$ and $\frac{\partial F}{\partial W_2}$ be (well) defined?

Jan Stuller
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1 Answers1

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For question 1, you are correct that $F$ is not differentiable with respect to $t$. It's not exactly clear what it would mean for $F$ to be differentiable with respect to $W_t$ because in general we don't think of $W_t$ changing without $t$ also changing. So while $F$ is clearly differentiable with respect to its first argument, I'm not sure that's exactly what you want when you talk about $F(W_t,t)$ being differentiable with respect to $W_t$.

For the second question, which also addresses question 3, we have something very similar to the fundamental theorem of calculus: If $H$ is adapted, right-continuous, and bounded then $$\lim_{t \rightarrow 0} \frac{1}{W_t} \int_0^t H_s dW_s = H_0$$ in probability.

Proof: Fix $K, \varepsilon > 0$. We compute

\begin{align*} &\phantom{= }P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right)\\ &= P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon \right) \\ &= P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) + P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} \le K \right) \\ &\le P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) + P\left(\frac{|W_t|}{\sqrt{t}} \le K \right). \end{align*}

Now we focus on the first term and use Markov's inequality:

\begin{align*} P\left( |\frac 1{W_t}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) &\le P\left( |\frac {1}{K\sqrt{t}}\int_0^t (H_s - H_0)dW_s| > \varepsilon, \frac{|W_t|}{\sqrt{t}} > K \right) \\ &\le P\left( |\frac {1}{K\sqrt{t}}\int_0^t (H_s - H_0)dW_s| > \varepsilon \right) \\ &\le \frac{1}{\varepsilon^2}\mathbb{E}\left[ \frac{1}{K^2 t} |\int_0^t (H_s - H_0)dW_s|^2\right] \\ &= \frac{1}{\varepsilon^2K^2 t} \mathbb{E}\left[\int_0^t |H_s-H_0|^2 ds \right] \\ &\le \frac{1}{\varepsilon^2K^2 t} \mathbb{E}\left[ t \sup_{0 \le s \le t} |H_s-H_0|^2 \right] \\ &= \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right]. \end{align*}

For the second term, by the scaling properties of Brownian motion we have $P\left(\frac{|W_t|}{\sqrt{t}} \le K \right) = P(|X| \le K)$ where $X$ is a standard normal random variable. Combining everything, we have

\begin{align*} \lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) &\le \lim_{t \rightarrow 0} \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right] + P(|X| \le K). \end{align*}

Since $H$ is bounded, $\sup_{0 \le s \le t} |H_s-H_0|^2$ is also bounded, and since $H$ is right continuous we know that $\lim_{t \rightarrow 0} \sup_{0 \le s \le t} |H_s-H_0|^2 = 0$ almost surely, so by the dominated convergence theorem we have

\begin{align*} \lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) &\le \lim_{t \rightarrow 0} \frac{1}{\varepsilon^2 K^2}\mathbb{E}\left[\sup_{0 \le s \le t} |H_s-H_0|^2 \right] + P(|X| \le K) \\ &= P(|X| \le K). \end{align*}

Finally, sending $K \rightarrow 0$ gives $$\lim_{t \rightarrow 0} P\left( |\frac 1{W_t}\int_0^t H_s dW_s - H_0| > \varepsilon \right) = 0.$$

This result is actually still true if $H$ is continuous but unbounded by localizing $H$ first, so that would be the result you need for question 3. Let me know if you want me to add on the details for that.

user6247850
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    Thank you so much. I guess what still confuses me regarding question 1 and your answer that $F(W_t,t)$ is not differentiable w.r.t. "$t$" is that I could just define $$F(W_t,t):=0.5W_t^2-0.5t$$ and then apply Ito's lemma to this $F(W_t,t)$, which would involve computing $\frac{\partial F}{\partial t}$. So it seems that if we define $$F(W_t,t):=\int_0^tW_hdWh$$ we cannot carry out the derivative $\frac{\partial F}{\partial t}$, but we may carry out that derivative on the "result" (which "equals" the definition). If the latter equals the former, then we can only differentiate the former? – Jan Stuller Dec 30 '20 at 23:21
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    Ito's lemma involves computing the partials of $F(x,y) := \frac 12(x^2-y)$, and then evaluating those partials at $x=W_t$ and $y=t$. The function $\frac 12 (W_t^2-t)$ is not differentiable with respect to $t$, but Ito's lemma lets us write it in terms of Ito and Lebesgue integrals. The main problem with applying Ito to the first definition $F(W_t,t) := \int_0^t W_h dW_h$ is that it is not obvious that the right hand side is just a function of $W_t$ rather than the whole path $W_s, s \le t$, which is why rewriting it as $\frac 12 (W_t^2-t)$ is helpful. – user6247850 Dec 30 '20 at 23:29
  • In all honesty, what inspired me to ask this question was another question on math SE here: I was hoping that your answer would allow me to "solve" that question, but I still wouldn't know how to answer it. So perhaps this is somewhat diverging from the original question, but is there a way you could tie your answer into showing how $$\int_{h=0}^{h=t}W_1(h)dW_2(h)$$ can be differentiated with respect to $W_1$ and $W_2$ in practice? (I suppose this would qualify as an explicit answer to my 3rd question). – Jan Stuller Dec 31 '20 at 10:57
  • Alternatively, feel free to answer the other question from the other user, that would also be super helpful :) – Jan Stuller Dec 31 '20 at 11:04
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    I answered the other question, and hopefully that helps in clarifying why there are problems in talking about differentiating this process. The point of Ito's formula is to write something as a sum of Ito and Lebesgue integrals, so if the process is already an Ito integral then Ito's formula doesn't do much. – user6247850 Dec 31 '20 at 15:48
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    What in your opinion is the main advantage of being able to rewrite a process as a sum of a Riemann and and Ito integral, using Ito's lemma? I am asking because I have seen probably more examples going in the "other direction", i.e. to use Ito's lemma to solve a stcohastic integral. A classical example would be $$\int_{h=0}^{h=t}W_hdW_h$$ which can, of course, be integrated via first principles, but the easiest way is to apply Ito's lemma to $W_t^2$. Well, I suppose that sort of answers my own question: being able to rewrite a process as a sum of Riemann + Ito can help us solve Ito integrals? – Jan Stuller Jan 01 '21 at 14:05
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    That is definitely one advantage of Ito's lemma. Another is that it tells us that if $f$ is sufficiently regular and $X_t$ is a semi-martingale, then $f(X_t)$ is a semi-martingale. It also decomposes $f(X_t)$ into a (local) martingale part and a drift term, so we can see when $f(X_t)$ is a local martingale, i.e. when the drift term is $0$. There are of course other advantages, but those are a few of the big ones that jump out immediately. – user6247850 Jan 01 '21 at 16:24