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Let $X_t = tW_{\frac{1}{t}} \forall t>0$ and $X_0 = 0$. I am trying to show that this process is a brownian motion under some measure P.

I have shown that it is continuous and that it is distributed normally with mean zero and variance t. However, I am trying to show that it has normally distributed increments:

let $t>s\ge0$ then I want to show that $X_{t+s}-X_s \sim N(0,t)$. Showing that this has mean zero and variance t is easy. However, the solution I have states that since this is a sum of gaussian random variables, then it must also be gaussian. However, when looking online I can see that the sum of two normal rv's need not be gaussian, that is, those rvs must be jointly gaussian, which then implies that the sum is gaussian. Am I missing something, or is the solution skipping over an important distinction?

WeakLearner
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  • Since $(W_t){t\ge0}$ is a Gaussian process, $(X{t+s},X_s)$ is jointly Gaussian. – Ian Mar 09 '15 at 09:09
  • @Ian so there is a distinction there between gaussian rv's and gaussian processes? – WeakLearner Mar 09 '15 at 09:10
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    By definition of a Gaussian process, for any $t_1<\cdots<t_n$, $(W_{t_1},\ldots,W_{t_n})$ is jointly Gaussian. Does this answer your question? – Ian Mar 09 '15 at 09:11
  • @Ian So I understand that Wt is a gaussian process by definition, but can we make the assumption that Xt is a gaussian process simply as it is a multiple of a gaussian process? – WeakLearner Mar 14 '15 at 02:49
  • It is not an assumption to be made, but rather a true statement! Indeed, a (finite) linear combination of samples $(X_{t_1},\ldots,X_{t_n})$ is clearly a linear combination of $(W_{s_1},\ldots,W_{s_n})$ which is a Gaussian random variable since $(W_t){t\ge0}$ is a Gaussian process. This proves that $(X_t){t\ge0}$ is a Gaussian process. – Ian Mar 15 '15 at 23:05

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