I am currently studying for an exam, and in studying one of the examples I am trying to construct is a bounded simple process $H_t$ that changes value only once such that$$\int_0^t H_s\,dB_s$$does not have a normal distribution. However, I was not able to come up with anything. Could someone please help?

EDIT: Progress so far. How about any $H_s = \textbf{1}_{s < T}$, with $T$ random? But I do not know how to go about showing that this works...

  • $\begingroup$ Please add the [self-study] tag & read its wiki. $\endgroup$ Sep 18, 2015 at 15:37
  • $\begingroup$ What about $T = \inf\{s|B_s>K\}$ with $K$ real? $\endgroup$
    – RUser4512
    Sep 18, 2015 at 15:44

2 Answers 2


If $T$ is allowed to depend on $B$, you could use $T_K=\inf\{t|B_t>K\}$. Your indicator becomes equivalent to $H_s=1_{\sup_{u<s}B_u<K}$.

The integral itself becomes:$$\int_0^t H_s\,dB_s=B_{\min(t,T_K)}$$

From where it is easy to conclude.


For a non-random $T$ you can split the integral and get a sum of two Gaussians. This is clearly Gaussian.

For a random $T$, say with finite support, you can condition on $T$ to get the previous result. Marginalizing on $T$, you will be getting a mixture of Gaussian.


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