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Given a continuous stochastic process with respect to time with finite variance at a given $t$. Does it necessarily imply, as $d\to 0$,

1) the covariance between $t$ and $t+d$ approaches the variance at $t$?

2) when both variance at $t$ and $t+d$ are positive, the correlation coefficient between $t$ and $t+d$ approaches $1$?

3) if the answer to 2) is in the negative, that the correlation between $t$ and $t+d$ can not approach $0$?

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No to questions 1 and 2.

Let $X_t$ be the process defined in this answer, which is continuous, satisfies $E X_t = 0$ for $t < 1$, and for $t \ge 1$ we have $X_t = 1$ almost surely. Let $U$ be any random variable with nonzero finite variance, and independent of the process $\{X_t\}$. Set $V_t = U X_t$, which is also continuous. Since $X_t$ and $U$ are nonconstant, $\operatorname{Var}(V_t) > 0$ for all $t$. But for $t < 1$ we have $$ \begin{align*} \operatorname{Cov}(V_t, V_1) &= E[V_t V_1] - E[V_t] E[V_1] \\ &= E[X_t U^2] - E[X_t U] E[U] \\ &= E[X_t] E[U^2] - E[X_t] E[U]^2 && \text{by independence} \\ &= 0 \end{align*} $$ since $E[X_t] = 0$. This resolves your questions 1 and 2 in the negative.

For question 3, I don't have an example off the top of my head, but I would probably try a process like $W_t = (X_t+1)U$.

Generally, the continuity of a stochastic process is a statement about almost sure convergence, and you really should not expect almost sure convergence to imply anything about convergence of moments.

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  • $\begingroup$ Ingenious trick of stopping time and multiplication with an independent random variable. Actually, your counterexample for questions 1) and 2) is exactly one for question 3) since its correlation coefficient is zero. Do you agree? Thank you. $\endgroup$
    – Hans
    Feb 8 '15 at 22:13

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