$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?

  • 3
    $\begingroup$ Yes you are. You can also use the property that for any $L^2$ function $f$, $\int_0^T f(t) dB(t)$ is normally distributed with mean zero and variance $\int_0^T f(t)^2 dt$ $\endgroup$ – Antoine Conze May 22 '18 at 11:15

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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