# Is it meaningful to look at $\int f(W_t, t) \,dt$?

CONTEXT (can skip):

My textbook looks at two things -

1) Ito integrals for deterministic functions—i.e. $$\int f(t) \,dW_t$$. We are able to say that they are normally distributed, with a mean of 0 and a variance of $$\int f^2(t) \,dt$$.

2) Ito integrals for stochastic functions—i.e. $$\int f(W_t, t) \,dW_t$$. We aren't able to find their distribution in general; but we can conclude that they have a mean of 0, and ito isometry can be used to get a neat expression for their variance.

QUESTION:

Is it meaningful to look at something of the form $$\int f(W_t, t) \,dt$$?

I feel the answer is yes. We won't be able to find the distribution of $$\int f(W_t, t) \,dt$$ in general, but we can say that it will have a mean of $$\int E[f(W_t, t)] \,dt$$. I also think that the second order raw moment $$E[(\int f(W_t, t) \,dt)^2]$$ will be given by $$\int \int E[f(W_t, t) f(W_s, s)] \,dtds$$, which can then be used to find the variance.
• Such time integrals are useful. The simplest case is obviously $f(t,W_t)=W_t$ in which case we know the distribution of $\int f(t,W_t)\mathrm{d}t$. In general, a time integral is closely linked to Asian options which are written on the average of the underlying asset price. You could also let $f$ be the price of an option and look at the term structure of options as opposed to the popular integral across the moneyness of options (for static hedges). – Kevin Feb 28 '20 at 7:29
• (1) I can't see what the distribution would be when $f(W_t, t) = W_t$; some elaboration would be very helpful. (2) The point regarding Asian options was insightful. (3) Are the expressions for the mean and variance of $\int f(W_t, t) \,dt$ stated by me in the question details correct? Thank you for your help! – Dhruv Gupta Feb 28 '20 at 7:36