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Let $W(t)$ be a standard Brownian motion (BM), in particular $W(t) \sim \mathcal{N}(0,t)$. Then it is easily shown that $\int_0^T W(t) dt \sim \mathcal{N}(0, T^3/3)$.

Question: What is the distribution of $X := W(T) + \int_0^T W(t) dt$? Are these two random variables not dependent? Since they are, how do we know $X$ is normal? I've always thought that the sum of two normals is not necessarily normal, especially when they are correlated. Thanks!

bcf
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1 Answers1

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As a linear combination of a centered gaussian process, the random variable $X$ is centered normal and its distribution is fully characterized by its variance $\sigma^2_T=E(X^2)$, where $$ \sigma^2_T=E(W(T)^2)+2\int_0^TE(W(T)W(t))\mathrm dt+2\int_0^T\!\!\!\int_0^t E(W(t)W(s))\mathrm dt\mathrm ds, $$ which should be evaluated easily. Alternatively, $$ X=\int_0^T W(t)\mathrm d\mu_T(t), $$ for some measure $\mu_T$ one can write down explicitely, hence $$ \sigma^2_T=\iint_{[0,T]^2}\min\{t,s\}\,\mathrm d\mu_T(t)\mathrm d\mu_T(s). $$

Did
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  • Ok, thanks for that. I think the point of confusion for me is whether this particular sum of normals is indeed normal, and it seems you've concluded that it is based on each summand being centered. I've always understood that unless the summands are uncorrelated, their sum may not be normal. What am I forgetting? :/ – bcf May 30 '14 at 17:32
  • Every linear combination of a normal family is gaussian. Recall that a family being normal is much more restrictive than each member of the family being normal. – Did May 30 '14 at 17:42
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    @bcf If we set $t_j^n := T \cdot j/n$ and define $$X^n := W(T) + \sum_{j=0}^{n-1} W(t_{j+1}^n) \cdot \left( t_{j+1}^n-t_j^n\right)$$ then $X^n \to X$ almost surely (and in $L^2$). As $(W_{t_1^n},\ldots,W_{t_n^n})$ is (jointly!) Gaussian, we know that $X^n$ is Gaussian (as a linear combination of jointly gaussian RVs). Hence, $X$ is Gaussian. – saz May 30 '14 at 18:34
  • @saz Im wondering when you can decide if a sum of gaussians is made up of jointly gaussian r.v. Is independence a sufficient condition? And as above is independent increments another sufficient condition? – JKnecht Oct 09 '19 at 22:00
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    @JKnecht Yes, exactly. – saz Oct 10 '19 at 05:42