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If $W(t)$ is a Wiener process and $V(t) = t\cdot W(1/t)$ is it possible to say that

Since $W(1/t)\space \sim N(0,1/t)$

that $V(t) \sim t\cdot N(0,1/t)$?

And if so then is $t\cdot N(0,1/t) = N(0,t^2\cdot 1/t) = N(0,t)$?


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Please don't use $*$ for multiplication. In mathematics, that symbol is commonly used for convolution. – Harald Hanche-Olsen Oct 10 '12 at 16:47
up vote 1 down vote accepted

Yes. In fact $t\mapsto t\cdot W(1/t)$ is itself a Wiener process, strange as it may seem.

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Thanks. Is $V(t) \tilde{}N(0,t)$ then? – Dirk Calloway Oct 10 '12 at 17:00
Yes, that comes from the scaling behaviour of Gaussian variables. Recall that the second argument in $N(0,t)$ is the variance, and that the variance is a quadratic function in the variable itself. – Harald Hanche-Olsen Oct 10 '12 at 17:03

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