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If a stochastic process $X_{t}$ $\sim N(0,t^3/3)$ and $Y_{t}$ is defined as follows:

$Y_{t} = X_{t}/t$, if $t>0$ and $ Y_{t} = 0$, if $t=0$, then how can I show that $Y_{t}$ is continuous in $t>=0$ almost everywhere?

I started as follows: $X_{t}$ is continuous in $t$ and since the probability measure of $w$ such that $Y_{0}(w)=0$ is $0$, $Y_{t}$ is continuous in $t$ almost everywhere. Any help would be great.

Thanks Trambak

Later. (Copied from an answer where Trambak mistakenly added it —Mariano)

Well $X_t$ is actually $\int_{0}^{t}(t-s)dW_{s}$, where $W_{s}$ is a Brownian Motion. One can show that $X_{t}$ will have a distribution as given in the question above. From this point onwards, how can I prove that $Y_{t}$ is continuous in $t$ with probability 1?

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Is $X_t$ supposed to be continuous? – Ilya Jan 29 '13 at 12:18

Hint: Rewrite $\int_{0}^{t}(t-s)dW_{s}$ as $\int^t_0 W_s ds$.

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