Using Itô’s Lemma to Evaluate an Expected Brownian Path Integral

Summary

The document considers the expectation of a time integral whose integrand is a bounded function of Brownian motion. It applies Itô’s lemma to the arctangent of the process, rewriting the ordinary time integral as a stochastic integral minus a terminal-value term. Under the usual integrability conditions, the stochastic integral has zero expectation, so the method reduces the calculation to the expected arctangent of the Brownian endpoint.

The answer claims a specific value after setting the endpoint to a fixed value. That step requires care: conditioning Brownian motion on its endpoint changes the expectation of the stochastic integral, and a fixed endpoint alone does not specify the distribution of the path. Thus the stated result does not follow from the preceding unconditional zero-mean argument if the question means conditional expectation given the endpoint. The Ito-lemma transformation is useful, but the conditioning and initial-value assumptions must be made explicit.

Key ideas

• Itô’s lemma can rewrite certain ordinary time integrals along Brownian paths using a stochastic integral and a terminal function value.
• A square-integrable Itô integral has zero expectation under the standard unconditional setup.
• Conditioning on a Brownian endpoint changes the path distribution and can change the expected stochastic integral.
• The endpoint and initial-value assumptions must be clear before assigning a numerical expectation.

Tags

# How to compute the expectation of integral of this random function?


# How to compute the expectation of integral of this random function?












Let $W_t$ be a standard wiener process and

$$Y_t=\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds$$

If $W(t_0)=\sqrt{3}$, then how can we compute $\mathbb{E}[Y(t_0)]$?

Is $\mathbb{E}[Y(t_0)]=0$?

## Answer by user16651 (score 9, accepted)

https://quant.stackexchange.com/a/28010

1. This integral is not Ito's Integral. Indeed $Y_t$ is a random time change with time change rate $\frac{W_t}{1+W_t^2}.$ (Oksendal, Sixth edition,page 147)

2. Sometimes this trick is useful.Indeed we assume that we are going to solve Riemann integral !.

Let $$f''(x)=\frac{-2x}{(1+x^2)^2}$$ then $$f'(x)=\left(\frac{1}{1+x^2}\right)+c_1$$ and $$f(x)=\tan^{-1}(x)+c_1x+c_2$$ set $c_1=c_2=0$.By application of Ito's lemma we have $$f(W_t)=f(W_0)+\int_{0}^{t}f'(W_s)dW_s+\frac{1}{2}\int_{0}^{t}f''(W_s)ds$$ therefore $$\tan^{-1}(W_t)=\int_{0}^{t}\frac{1}{1+W_s^2}dW_s-\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds$$ in other words $$\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds=\int_{0}^{t}\frac{1}{1+W_s^2}dW_s-\tan^{-1}(W_t)$$ thus $$\mathbb{E}\left[\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds\right]=\underbrace{\mathbb{E}\left[\int_{0}^{t}\frac{1}{1+W_s^2}dW_s\right]}_{0}-\mathbb{E}[\tan^{-1}(W_t)]$$ as a result $$\mathbb{E}\left[\int_{0}^{t}\frac{W_s}{(1+W_s^2)^2}ds\right]=-\mathbb{E}[\tan^{-1}(W_t)]$$ Now set $t=t_0$ $$\mathbb{E}\left[\int_{0}^{t_0}\frac{W_s}{(1+W_s^2)^2}ds\right]=-\mathbb{E}[\tan^{-1}(W_{t_0})]=-\mathbb{E}[\tan^{-1}(\sqrt{3})]=-\frac{\pi}{3}$$ finally

> $$ \color{red}{\mathbb{E}[Y_{t_0}]=-\frac{\pi}{3}}$$

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