Deriving $E[(B_t-x)^2] = nt$ in Brownian motion

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I am studying Stochastic DE with Oksendal's textbook. I am completely new to this topic, and I am having some problems going through it, which is sad.

I get for a certain variable to have a normal distribution, that variable needs to have $$p(t,x,y) = (2\pi t)^{-n/2}\exp(-|x-y|^2/2t)$$

And, making its multidimensional version and driving its characteristic equation results in the fact that the probability density function for multi-normally distributed variables needs to have the characteristic equations with :$M = E^x[Z]$ and $c_{jm} = E^x[(Z_j - M_j)(Z_m - M_m)]$ --image: MandCjm

What I do not get is how did you suddenly arrive at these, the expected values for $B_t$ and $(B_t-x)^2$:$E^x[B_t]=x$ and $E^x[(B_t-x)^2]=nt$ -- image:expvalues.

There must be a really simple step in-between, but I can't get it and it bothers me a ton. Also, Would you be kind to explain to me how did we get the covariance matrix C? That $I_n$ must be some kind of indicator function, but, frankly, I do not know what that represents.

btw, Sorry for using images instead of writing the formulas out--I am not yet used to write formulas in LaTeX.