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Let $\Omega = [0,1] $, $ \mathcal{F}$ - $ \sigma$-algebra and $ P$ - uniform distribution on $(\Omega, \mathcal{F})$. For $t \in [0, \infty)$ we define $X_t(\omega) = \omega t$. What is the two-dimensional probability distribution? Is it that we assume $x_2 > x_1$ and calculate $ P(X_t < x_1, X_t < x_2 )$?

$ P(X_t < x_1, X_t < x_2 ) \\ = P( X_t < x_1)P(X_t < x_2 | X_t < x_1) \\ = P(X_t < x_1) P(X_t < x_2 | X_t < x_1 \land x_1 < x_2) \\ = P(X_t < x_1) \cdot 1 = \frac{\omega t}{x_1}$

What is the n-dimensional probability distribution? Is it the same value?

M_T
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  • You need to take different values of $t$ for different values of $x$, i.e. find $P(X_{t_1}\le x_1,X_{t_2}\le x_2)$ where $t_1<t_2$ (but no assumptions are made on $x_1$ and $x_2$). – Jason Dec 06 '17 at 22:29
  • Then it's $P(X_{t_1} \leq x_1, X_{t_2} \leq x_2) = \frac{\omega t_1}{x_1} \frac{\omega t_2}{x_2}$ right? And n-dimensional probability is $P(...) = \omega^{n} \underset{i}\Pi \frac{t_i}{x_i}$ – M_T Dec 06 '17 at 22:37
  • No, the random variables are certainly not independent. – Jason Dec 06 '17 at 23:46
  • Is $ P (X_{t_1} \leq x_1 , X_{t_2} \leq x_2 ) = P (X_{t_1} \leq x_1 , X_{t_2} - X_{t_1} \leq x_2 - x_1 ) = \frac{\omega t_1}{x_1} \frac {\omega (t_2 - t_1)}{x_2 - x_1}$ ? – M_T Dec 08 '17 at 19:16
  • No. Neither equality you wrote there is correct. Go back to the definition of $X_t$ and use similar steps to what you did in your original argument. – Jason Dec 08 '17 at 23:04
  • If $X_{t_1} = \omega t_1 $ and $ X_{t_2} = \omega t_2 $ then $X_{t_2} = X_{t_1} \frac{t_2}{t_1}$, right? And then I should calculate $ P(X_{t_1} \leq x_1, X_{t_1} \leq \frac{t_1}{t_2} x_2) $? – M_T Dec 08 '17 at 23:33

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