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How should I interoperate $X_{t_j}$, does usually when I previously have seen double subscript it refer to having multidimensions.

In this case it looks like we are splitting up t into sections. In other words we could have that t goes from 0 to T, in steps of t.

however, is the second subscript referring to splitting up [0,t] into furter increments?

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ALEXANDER
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    $X_t$ is a (random) function of $t.$ $X_{t_i}$ is that function evaluated at $t=t_i.$ The $t_i$ are an evenly spaced sequence of sample points in the interval... compare to the evaluation points for a Riemann integral or something like that. – spaceisdarkgreen Jul 16 '18 at 23:51
  • @spaceisdarkgreen So I could consider it being the ith element of $X_t$ ? – ALEXANDER Jul 17 '18 at 00:49
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    No... that's not what I said. – spaceisdarkgreen Jul 17 '18 at 01:06
  • @spaceisdarkgreen Would you be able to write more in depth answer to the solution provided to this answer, where he uses the term ith element.: https://math.stackexchange.com/questions/2710012/proving-exb-s-b-t4-nn2t-s2 – ALEXANDER Jul 17 '18 at 02:06
  • They are referring to the $i$-th component of an $n$-dimensional Brownian motion which is a completely different concept. Their solution looks right (and is an elaboration of Did's suggestion in the comments) and I don't think I could improve on it. – spaceisdarkgreen Jul 17 '18 at 02:44
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    In this example in this question, the $i$ refers to the $i$-th time step in a discretization of the time axis. In other words we're taking a function defined on some interval, but only looking at a finite number of points (and then taking a limit as we go to finer and finer spaced points in order to recover information about the continuous time function... as we do in, say, the Riemann or Stieljes integral) – spaceisdarkgreen Jul 17 '18 at 02:55
  • @spaceisdarkgreen Thank you for taking the time, this really helps me understand what is going on. I have one more question, in the case of the multi dimensional Brownian motion, what I dont get is: If I look at $B_t$, and we say we have discrete number of case $B_t$={$x_1$$,x_2$ to whatever number of cases there are} However, from that question what I see is I have B_t and B_s, now does that mean that s and t are to different dimensions. – ALEXANDER Jul 17 '18 at 03:45
  • In other words I dont see why the ith subject refer to ith dimension, is it simply that $B_s^1$ ={vector of random numbers} and $B_s^2$={another vector of random numbers} – ALEXANDER Jul 17 '18 at 03:45
  • $s$ and $t$ are different times. You can view the time as a dimension if you wish, but if so it is a completely orthogonal dimension to the components of $B.$ Note the other problem makes no mention or use of discrete time, so the time index is continuous there... so $B^1_s$ is a vector of sorts, but it has an uncountably infinite number of components. – spaceisdarkgreen Jul 17 '18 at 03:53
  • @spaceisdarkgreen, Yes, that make sense, so in other words $B_s$ is simply $B(s)$ and $B(s)^i$ is used if you have several Brownian motions. $B_s^i$ where Bs1 is a vector of sorts, but it has an uncountably infinite number of components. and Bs2 is another vector of sorts? – ALEXANDER Jul 17 '18 at 03:59
  • Yes, but a simpler thing to say, more in keeping with the usual way this is talked about is that each of the components $B^i_t$ is a function of time (or a function on some interval), or that the whole thing B_t is a vector-valued function of time. The individual components are all brownian motions, sometimes correlated (though they are assumed independent in that other question). – spaceisdarkgreen Jul 17 '18 at 04:04
  • @spaceisdarkgreen Perfect, I need to be carefull when I read Bernt Oksendals book, as he uses $X_{t_i}$ for both discrete increment and multidimensional case. Obviously with more explanation, but I got tricked to think that the subscript earlier in the book which was for the multidimensional case was still the same as when he uses the subscript in this question I have posted here. – ALEXANDER Jul 17 '18 at 04:12

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