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When we want to obtain the order of $\int_{0}^{T} B_{t} d t$, we can use the scale property of Brownian motion.

Let $B$ be a Brownian motion. Is the order of $\int_{0}^{T} B_{t} d t$ correctly calculated:

$$\int_{0}^{T} B_{t} d t=\int_{0}^{1} B_{s T} \cdot T d s=\int_{0}^{1} T^{1/2} \cdot T B_{s} d s=T^{1+1/2} \int_{0}^{1} B_{s} d s$$

Alex
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user381975
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  • What does "order of a stochastic integral" mean? – Pontus Hultkrantz Dec 19 '20 at 22:20
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    @PontusHultkrantz Seems to be related to the Landau symbol. Consider the continuous function $F(T) = \int_0^T B_t\text{d}t$. What is the order of that function (= how quickly does it grow in $T$)? – Kevin Dec 19 '20 at 22:40
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    As @fesman says, the problem is when you use $B_{sT}=\sqrt{T}B_s$. You refer to this scaling property. But all it states is that both processes $B_{sT}$ and $\sqrt{T}B_s$ have the same distribution (are both Brownian motions). They are not equal for each realisation. So, for a fixed $T$, the random variables $\int_0^T B_t\text{d}t$ and $T^{3/2}\int_0^1 B_s\text{d}s$ are only equal in distribution (have the same moments) but don't necessarily have the same value if you consider one fixed outcome $\omega\in\Omega$. – Kevin Dec 20 '20 at 08:36
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    For example, consider mean and variance. By Fubini, both integrals are clearly zero. But consider second moments. From @Gordon, we know $\text{Var}\left[\int_0^T B_t\text{d}t\right]=\frac{1}{3}T^3$ and using your result, that $\int_0^T B_t\text{d}t\overset{d}{=}T^{3/2}\int_0^1 B_t\text{d}t$, we also get $\text{Var}\left[T^{3/2}\int_0^1 B_s\text{d}s\right]=T^3\text{Var}\left[\int_0^1 B_s\text{d}s\right]= \frac{1}{3}T^3$ – Kevin Dec 20 '20 at 08:44
  • @user381975 Regarding your question in the comments, what exactly do you mean by this? – fes Dec 20 '20 at 11:21
  • Thanks, I mean both processes $\int_{0}^{T} B_{t} d t$ and $T^{1+1/2} \int_{0}^{1} B_{s} d s$ have the same distribution. – user381975 Dec 20 '20 at 15:13

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This (in particular the 2nd equality) is incorrect. Let $\omega \in \Omega$ denote the sample realization. You state:

$$\int_{0}^{T} B_{t} (\omega) d t=T^{1+1/2} \int_{0}^{1} B_{s}(\omega) d s$$

For example assume $T>1$. This is claiming that in order to compute the integral you only need to consider realizations of $B_t$ between $[0,1]$ and then scale them up. However, the value of this integral also depends on the random realizations of $B_t$ between $[1,T]$, which cannot be deduced simply from its values between $[0,1]$. The LHS can take a value clearly different from the RHS depending on $\omega$. Intuitively if $W_t$ is e.g. a stock price, the integral represents a type of weighted average of the stock price over $[0,T]$. But you cannot calculate this average merely based on the stock price over $[0,1]$.

The probability that $\int_{1}^{T} B_{s}(\omega) d s$ takes any particular fixed value is zero (this integral is normally distributed). Hence the LHS and RHS are almost surely not equal. However, as @Kevin pointed out the LHS and RHS are equal in distribution that is these two random variables have the same moments.

fes
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