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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.

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    $\begingroup$ Just to make sure, is $\exp \big \{\frac{\theta^2}{\sigma}\big\}$ correct? $\endgroup$
    – Alecos Papadopoulos
    Commented Jan 2, 2014 at 18:46
  • $\begingroup$ @Alecos no, there is a typo, thanks for spotting it. $\endgroup$
    – Matteo Fasiolo
    Commented 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? $\endgroup$
    – Alecos Papadopoulos
    Commented Jan 2, 2014 at 18:51
  • $\begingroup$ Yes I noticed, I should remove "attention to details" from my CV. :) $\endgroup$
    – Matteo Fasiolo
    Commented Jan 2, 2014 at 18:54

1 Answer 1

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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",

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

DISCUSSION
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".

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    $\begingroup$ 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 ♦
    Commented Jan 2, 2014 at 19:23
  • $\begingroup$ @whuber I added some discussion. $\endgroup$
    – Alecos Papadopoulos
    Commented 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. $\endgroup$
    – Matteo Fasiolo
    Commented Jan 2, 2014 at 20:30
  • $\begingroup$ So what you are describing by $X(\theta)$ is a stochastic process with continuous and random index? $\endgroup$
    – Alecos Papadopoulos
    Commented 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 ♦
    Commented Jan 2, 2014 at 20:35

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