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jeq
jeq
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We can construct the probability density pY of Y by applying a bump to the probability distribution of X, $$ p_Y(x)=p_X(x)+\epsilon f(x_1)f(x_2)\cdots f(x_n). $$ Here, ε is a fixed real number and f: ℝ → ℝ is a continuous function of compact support and zero integral, $\int_{-\infty}^\infty f(x)\\,dx=0$$\int_{-\infty}^\infty f(x)\,dx=0$. Then, $\int_{-\infty}^\infty p_Y(x)\\,dx_k=\int_{-\infty}^\infty p_X(x)\\,dx_k$$\int_{-\infty}^\infty p_Y(x)\,dx_k=\int_{-\infty}^\infty p_X(x)\,dx_k$ for each k. So, the integral of pY over ℝn is 1 and, by choosing ε small, pY will be positive. Then it is a valid probability density function. Finally, as the integral along the kth direction (any k) agrees for pX and pY, the projection of X and Y onto ℝn-1 along the kth direction give the same distribution.

We can construct the probability density pY of Y by applying a bump to the probability distribution of X, $$ p_Y(x)=p_X(x)+\epsilon f(x_1)f(x_2)\cdots f(x_n). $$ Here, ε is a fixed real number and f: ℝ → ℝ is a continuous function of compact support and zero integral, $\int_{-\infty}^\infty f(x)\\,dx=0$. Then, $\int_{-\infty}^\infty p_Y(x)\\,dx_k=\int_{-\infty}^\infty p_X(x)\\,dx_k$ for each k. So, the integral of pY over ℝn is 1 and, by choosing ε small, pY will be positive. Then it is a valid probability density function. Finally, as the integral along the kth direction (any k) agrees for pX and pY, the projection of X and Y onto ℝn-1 along the kth direction give the same distribution.

We can construct the probability density pY of Y by applying a bump to the probability distribution of X, $$ p_Y(x)=p_X(x)+\epsilon f(x_1)f(x_2)\cdots f(x_n). $$ Here, ε is a fixed real number and f: ℝ → ℝ is a continuous function of compact support and zero integral, $\int_{-\infty}^\infty f(x)\,dx=0$. Then, $\int_{-\infty}^\infty p_Y(x)\,dx_k=\int_{-\infty}^\infty p_X(x)\,dx_k$ for each k. So, the integral of pY over ℝn is 1 and, by choosing ε small, pY will be positive. Then it is a valid probability density function. Finally, as the integral along the kth direction (any k) agrees for pX and pY, the projection of X and Y onto ℝn-1 along the kth direction give the same distribution.

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George Lowther
George Lowther
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No, it is not true that a process W satisfying the properties (1), (3) and (4) has to be a Brownian motion. We can construct a counter-example as follows.

This construction is rather contrived, and I don't know if there's any simple examples. Start with a standard Brownian motion W. The idea is to apply a small bump to its distribution while retaining the required properties. I will do this by first reducing it to the discrete-time case. So, choose a finite sequence of times 0 = t0 < t1 < ... < tn. Then define a piecewise linear process X by Xtk = Wtk (k = 0,1,...,n) and such that X is linearly interpolated across each of the intervals [tk-1,tk] and constant over [tn,∞).

Then, Y = W - X is a continuous process independent from X. In fact, Y is just a sequence of Brownian bridges across the intervals [tk-1,tk] and is a standard Brownian motion on [tn,∞). Also by linear interpolation, for any time t ≥ 0, Xt is a linear combination of at most two of the random variables Xt1,...,Xtn. The increments of W, $$ W_t-W_s = X_t-X_s + Y_t-Y_s, $$ are then a linear combination of at most 4 of the random variables Xt1,...,Xtn plus an independent term. So, choosing n ≥ 5, if it is possible to replace (Xt1,...,Xtn) by any other ℝn-valued random variable without changing the joint-distribution of any 4 elements, then the distributions of the increments Wt - Ws will be left unchanged. So, properties (1), (3), (4) will still be satisfied but the new process for W will not be a standard Brownian motion. It is possible to change the distribution in this way:

Let X = (X1,X2,...,Xn) be an ℝn-valued random variable with a continuous and strictly positive probability density pX: ℝn → ℝ. Then, there exists a random variable Y = (Y1,Y2,...,Yn) with a different distribution than X but for which the projection onto any n - 1 elements has the same distribution as for X.

That is, for any k1,k2,...,kn-1 in {1,...,n}, (Yk1,Yk2,...,Ykn-1) has the same distribution as (Xk1,Xk2,...,Xkn-1).

We can construct the probability density pY of Y by applying a bump to the probability distribution of X, $$ p_Y(x)=p_X(x)+\epsilon f(x_1)f(x_2)\cdots f(x_n). $$ Here, ε is a fixed real number and f: ℝ → ℝ is a continuous function of compact support and zero integral, $\int_{-\infty}^\infty f(x)\\,dx=0$. Then, $\int_{-\infty}^\infty p_Y(x)\\,dx_k=\int_{-\infty}^\infty p_X(x)\\,dx_k$ for each k. So, the integral of pY over ℝn is 1 and, by choosing ε small, pY will be positive. Then it is a valid probability density function. Finally, as the integral along the kth direction (any k) agrees for pX and pY, the projection of X and Y onto ℝn-1 along the kth direction give the same distribution.