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Consider an independent increment process $X_t$, such that $X_t$ follows a continuous distribution. Examples would be a Brownian motion, Gamma process or a stable Levy process.

When sampling from these processes on a regular time grid, linear interpolation in between time-grid points is sometimes used.

This got me thinking of the error committed to the statistics of $X_t$. That is with $X_t^{(1)} = (1-\lambda_t) X_{t_i} +\lambda_t X_{t_{i+1}}$ for $t_i < t < t_{i+1}$, with $\lambda_t = \dfrac{t-t_i}{t_{i+1}-t_i}$, how far can the distribution function of $X_t^{(1)}$ deviate from $X_t$ ?

I realize the question is vague, but I am trying to see if it can be made more precise, to ask what choice of $\lambda_t$ function would least distort the statistics.

Thanks for reading.

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Which information can be used? I mean, w.r.t. which sigma-algebra would you like to compare distributions of $X^{(1)}_t$ and $X_t$? –  Ilya Oct 17 '11 at 6:39
    
@Gortaur I think it is with respect to natural filtration of $X_t$. –  Sasha Oct 17 '11 at 15:36

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The formula you mention is based on the assumption that for every $r\leqslant t\leqslant s$, $$\mathrm E(X_t\mid \mathcal F^X_{\leqslant r}\vee\mathcal F^X_{\geqslant s})=\mathrm E(X_t\mid X_r,X_s)=X_r+\lambda_t(X_s-X_r). $$ The first equality holds for the Brownian motion (incidentally, this is the way Paul Lévy thought about Brownian motion) and for other processes with independent increments. For general independent increments processes, the best approximation of $X_t$ in terms of $(X_r,X_s)$ is $\mathrm E(X_t\mid X_r,X_s)$. Since (once again) the increments are independent, you are left with the task of estimating $$ \mathrm E(X_t-X_r\mid X_s-X_r)=\Phi_{r,s}(t;X_s-X_r) $$ and with using $X_r+\Phi_{r,s}(t;X_s-X_r)$ to approximate $X_t$.

By definition, $\Phi_{r,s}(r;x)=0$ and $\Phi_{r,s}(s;x)=x$. The case you mention is when, for every $r\leqslant t\leqslant s$ and every $x$, $\Phi_{r,s}(t;x)=x(t-r)/(s-r)$.

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Function $\Phi_{r,s}(t ; X_s-X_r)$ is the mean function of the bridge process, right ? These are easy to estimate. –  Sasha Oct 17 '11 at 13:19

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