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The author of this question might be more pleased with the following answer. Let $U$ be a uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + {\mathbf 1}(t=U)$, where ${\mathbf 1}$ denotes indicator function, is discontinuous at time $U$. However, for any choice of (fixed) times $t_i$, $i=1,...,n$, we have, almost surely, $W'(t_i) = W(t_i)$ for all $i$, and hence, trivially, $W'$ has the same distributional properties stated for $W$. Furthermore, if we define $W'$ by $W'(t) = W(t) + {\mathbf 1}(t \in UA)$, where $A$ is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$), then $W'$ is nowhere continuous (since $UA$ is dense in $(0,\infty)$); nevertheless, as before, almost surely $W'(t_i) = W(t_i)$ for all $i=1,...,n$ (since ${\rm P}(t \in UA) = {\rm P}(t/U \in A) = 0$).

Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2) The answer can be adapted to Lévy processes in general ($W$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions.

Finally, the author of this question wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text

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The author of this question might be more pleased with the following answer. Let $U$ be a uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + {\mathbf 1}(t=U)$, where {\mathbf 1} ${\mathbf 1}$ denotes indicator function, is discontinuous at time $U$. However, for any choice of (fixed) times $t_i$, $i=1,...,n$, we have, almost surely, $W'(t_i) = W(t_i)$ for all $i$, and hence, trivially, $W'$ has the same distributional properties stated for $W$. Furthermore, if we define $W'$ by $W'(t) = W(t) + {\mathbf 1}(t \in UA)$, where $A$ is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$), then $W'$ is nowhere continuous (since $UA$ is dense in $(0,\infty)$); nevertheless, as before, almost surely $W'(t_i) = W(t_i)$ for all $i=1,...,n$ (since ${\rm P}(t \in UA) = {\rm P}(t/U \in A) = 0$).

Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2) The answer can be adapted to Lévy processes in general ($W$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions.

Finally, the author of this question wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text

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The author of this question might be more pleased with the following answer. Let $U$ be a uniform(0,1) random variable, independent of a Brownian motion $W$. Then, the process $W'$ defined by $W'(t) = W(t) + 1(t=U)$, {\mathbf 1}(t=U)$, where${\mathbf 1$} denotes indicator function, is discontinuous at time$U$. However, for any choice of (fixed) times$t_i$,$i=1,...,n$, we have, almost surely,$W'(t_i) = W(t_i)$for all$i$, and hence, trivially,$W'$has the same distributional properties stated for$W$. Furthermore, if we define$W'$by$W'(t) = W(t) + 1(t {\mathbf 1}(t \in UA)$, where$A$is a dense set in $(0,\infty)$ of measure zero (and where $UA:= \{Ua: a \in A\}$), then$W'$is nowhere continuous (since$UA$is dense in$(0,\infty)$); nevertheless, as before, almost surely$W'(t_i) = W(t_i)$for all$i=1,...,n$(since$P(t {\rm P}(t \in UA) = P(t/U {\rm P}(t/U \in A) = 0$). Side notes: 1) Actually, as follows from the theory of Lévy processes, the almost sure continuity in the definition of Brownian motion is equivalent to almost sure cadlaguity (right-continuity with left limits); 2) The answer can be adapted to Lévy processes in general ($W\$ is a special case), showing that the almost sure cadlaguity in the definition of Lévy process is not implied by the other conditions.

Finally, the author of this question wanted to make sure that all the conditions are mutually independent.'' This is, however, not the case, if we split condition 3) into subconditions. See this thread: link text

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