Computing the Distribution of a Piecewise-Constant Itô Integral

Summary

The document works through an Itô integral whose integrand is constant on three successive time intervals. The integral is rewritten as a weighted sum of Brownian increments, making clear how each segment contributes a normally distributed random variable. Because the increments are independent and centered, the total is also normal with mean zero.

The accepted explanation verifies the variance using the Itô isometry: integrate the squared integrand over time. The intervals do not overlap, so their contributions add, yielding variance 47. This confirms the result proposed in the question. The calculation assumes standard Brownian motion and the stated deterministic integrand; it illustrates a basic distributional calculation rather than a general treatment of stochastic integrals with random or path-dependent integrands.

Key ideas

• A piecewise-constant Itô integral can be expressed as a weighted sum of Brownian increments.
• Independent Brownian increments make the resulting sum normally distributed.
• The integral has mean zero when its deterministic integrand is integrated against standard Brownian motion.
• The variance equals the time integral of the squared integrand, which is 47 in this example.

Tags

# Basic question on Ito integrals


# Basic question on Ito integrals












> $Let \space X(t) =\begin{cases} 2, \qquad\text{if} \space 0\le t \le 1 \\ 3, \qquad\text{if} \space 1 < t \le 3 \\ -5, \qquad\text{if}\space 3 < t \le 4 \end{cases} $ or in one forumala $X(t) = 2I_{[0,1]}(t)+3_{(1,3]}(t)-5_{(3,4]}(t)$. Give the $It\hat{o}$ integral $\int_0^4 X(t)dB(t)$ as a sum of random variables, give its distribution, specify the mean and the variance.

Here is what I tried:

$\int_0^4X(t)dB(t) = \int_0^12dB(t)+\int_1^33dB(t)+\int_3^4-5dB(t)$

$=2(B(1)−B(0)) + 3(B(3)−B(1))−5(B(4)−B(3))$

We can write this as the sum of Normal Random variables:

$\int_0^4X(t)dB(t) = 2N(0,1) + 3N(0,2)−5N(0,1)$

$= N(0,2^2) + N(0,2(3^2)) + N(0,(−5)^2)$

= $N(0,47)$

Thus, the $It\hat{o}$ integral $\int_0^4X(t)dB(t)$ is Normally distributed with mean $0$ and variance $47$.

Am I on the right track here?

## Answer by Bjørn Kjos-Hanssen (score 6)

https://quant.stackexchange.com/a/43189

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 terms $2_{[0,1]}(t), 3_{(1,3]}(t), 5_{(3,4]}(t)$ is identically 0, so the integral is just $$\int_0^4 2^2_{[0,1]}(t)+3^2_{(1,3]}(t)+5^2_{(3,4]}(t)\,dt=4(1-0)+9(3-1)+25(4-3)=47.$$

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