I have encountered the following integral: $$ \int_{- \infty}^{+ \infty} X(\theta) \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}d\theta $$ where, for fixed $\theta = \theta_0$, $X(\theta_0) \sim N(0, 1)$. $X(\theta_1)$ and $X(\theta_2)$ are independent if $\theta_1 \neq \theta_2$. Hence $X(\theta)$ is Gaussian white noise indexed by $\theta$.

I am not sure about how to interpret this expression. In particular, by thinking about the integral as the limit of a sum: $$ \frac{1}{N}\sum_{i=1}^N X(\theta_i), \;\;\;\;\;\;\; X(\theta_i) \sim N(0, 1) \;\;\;\;\text{and}\;\;\;\;\; \theta_i \sim N(0, \sigma) $$ for $i = 1, \dots, N$, I would say that it should converge to zero in probability, but I'm not completely sure.

  • 1
    $\begingroup$ Just to make sure, is $\exp \big \{\frac{\theta^2}{\sigma}\big\}$ correct? $\endgroup$ Jan 2, 2014 at 18:46
  • $\begingroup$ @Alecos no, there is a typo, thanks for spotting it. $\endgroup$ Jan 2, 2014 at 18:49
  • $\begingroup$ The "usual" $1/2$ is still missing, but I guess that's the way it should be here? $\endgroup$ Jan 2, 2014 at 18:51
  • $\begingroup$ Yes I noticed, I should remove "attention to details" from my CV. :) $\endgroup$ Jan 2, 2014 at 18:54

1 Answer 1


We have by assumption, $Y = X(\theta) \sim N(0,1)$. As corrected, the integral is, by the so-called "law of the unconscious statistician",

$$ \int_{- \infty}^{+ \infty} X(\theta) \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}d\theta = \int_{- \infty}^{+ \infty} X(\theta) f_{\theta}(\theta)d\theta = E(X(\theta))=E(Y) = 0$$

Responding to @whuber 's comment, and pending clarification from the OP, I understand $X(\theta)$ as a function of $\theta$, which in turn, by the information in the question, is a $N(0,\sigma)$ random variable. Then the density of $\theta$ is $$f_{\theta}(\theta) = \frac{\exp \big \{-\frac{\theta^2}{2\sigma^2} \big \}}{\sigma\sqrt{2 \pi}}$$ and the result I give above is immediate -but I may have misunderstood the information provided in the question regarding the meaning of "indexing".

  • 1
    $\begingroup$ I am having trouble making sense of the first two inequalities and the notation. According to the question, $X$ is a family of random variables indexed by $\theta$: it is a stochastic process. Thus the integral is not a Riemann or Lebesgue integral at all and does not appear to be an expectation: it would have to be a random variable. The meaning of $f_\theta(\theta)$ is mysterious. Could you please explain your interpretation, explain your notation, and justify your steps? $\endgroup$
    – whuber
    Jan 2, 2014 at 19:23
  • $\begingroup$ @whuber I added some discussion. $\endgroup$ Jan 2, 2014 at 19:46
  • $\begingroup$ I am quite sure that Alecos final result is right: the integral is equal to zero. On the other hand, as Whuber is saying, $X(\theta)$ is a stochastic process not a random variable, so I'm not sure about writing $Y = X(\theta)$. I used the notation $X(\theta)$ to mean that for every $\theta$ on the real line there is a random variable. I used the expression "indexed by" to convey the fact that $X(\theta_1), X(\theta_2), \dots$ are iid. $\endgroup$ Jan 2, 2014 at 20:30
  • $\begingroup$ So what you are describing by $X(\theta)$ is a stochastic process with continuous and random index? $\endgroup$ Jan 2, 2014 at 20:32
  • $\begingroup$ Matteo, if you are sure the answer is zero, then some part of your description of $X$ must be incorrect. Could you supply a link or reference to where this integral appeared so we could have some context for understanding it? $\endgroup$
    – whuber
    Jan 2, 2014 at 20:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.