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I have a question about the connection between continuous time white noise and discrete white noise (i.e. i.i.d gaussians). If I understand correctly you cannot derive a discrete white noise process from a continuous one, since the derivative of a Brownian motion (i.e continuous white noise) is too "rough".

Consider the following: Assume $X$ to be continuous time white noise and $f$ square Riemann integrable with $\int_0^1 f(t)^2 dt = 1$ then \begin{equation} \mathbb{E}\left[\left(\int_0^1 X(t) f(t) dt \right)^2 \right] = \int_0^1 f(t)^2 dt = 1. \end{equation}

On the other in the discrete case ($\epsilon_t$ i.i.d gaussian): \begin{equation} \mathbb{E}\left[\left(\frac{1}{n} \sum_{t=1}^n f(t/n) \epsilon_t \right)^2\right] = \frac{1}{n^2} \sum_{i=1}^n f(t/n)^2 \to 0. \end{equation}

I know this is mathematical quite imprecise. However, I wanted to know if I am misunderstanding something or if these quantities actually don't match up. If they don't: Is there an intuition why this is not the case?

FHG_12
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1 Answers1

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From my understanding, the former integral finds its usage in the engineering, but does not have a formal mathematical meaning. So you should not be surprised it may lead to contradictions, likely this is part of the reason why it is hard to give it a formal meaning. What does exist though is $$ \int_0^Tf(t)\mathrm dW(t) \tag{1} $$ where $W$ is a Brownian motion. For it we do indeed have $$ \Bbb E\left[\int_0^Tf(t)\mathrm dW(t)\right]^2 = \int_0^T f^2(t)\mathrm dt $$ which is called the Ito isometry. However, there is a trick here: you see $W(t)$ has a variance of $t$ so that $$ \Delta W(t) := W(t+\Delta t) - W(t) = X(t)\sqrt t $$ will imply that $X(t)$ is standard normal random variable, moreover $X(t)$ is independent from $X(t+\Delta t)$ since increments $\Delta W(t)$ are independent from each other by the definition of the Brownian motion. So that integral in $(1)$ should be more like $$ \int_0^Tf(t)X(t)\sqrt{\mathrm dt} \tag{2} $$ rather than what you had. Notice the square root of the differential, and try to derive something of that kind for your second formal where you have the sum. Note thought that $(2)$ does not have a formal mathematical meaning either, but it is inspired by $(1)$ which does, and may give a hint about which sum to consider to get the desired result.

SBF
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