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Assuming that $\{ W ( t ) | t \geq 0 \}$ is a Brownian motion, I'm trying to determine the distribution of the random variable $W ( 1 / 2 ) - 3 W ( 4 )$.

Here is my try: From properties of Wiener process, we know that

$$W ( 0 ) = 0$$

I can then write that \begin{align*} W ( 1 / 2 ) - 3 W ( 4 ) &= \left( W ( 1 / 2 )-W(0)\right) - \left(3 W ( 4 )-3W(0)\right)\\ &= \mathcal{N}(0,1/2)-3\mathcal{N}(0,4)\\ &= \frac { 1 } { \sqrt { 2 \pi \frac{ 1 }{ 2 } } } \exp \left( - \frac { ( x ) ^ { 2 } } { 2 \times \frac{ 1 }{ 2 } } \right)-\frac { 3 } { \sqrt { 2 \pi 4 } } \exp \left({ - \frac { ( x ) ^ { 2 } } { 2 \times 4 } }\right)\\ &=\frac { 1 } { \sqrt { \pi } } \exp \left( - ( x ) ^ { 2 } \right) - \frac { 3 } { \sqrt { 8 \pi } } \exp \left( - \frac { ( x ) ^ { 2 } } { 8 } \right) \end{align*}

First, I was wondering if this approach is correct, and second, if there is a better way to represent it as a distribution.

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There are a number of errors in what you have done. There is confusion between distribution function and density. There is a wrong assumption that density of $X-Y$ is density of $X$ minus density of $Y$. More importanlty the distribution function of the difference of two random variables cannot be determined by their individual distributions without some independence assumption.

The correct solution is as follows: $W(\frac 1 2) -3 W(4)=-3(W(4)-W(\frac 1 2))-2W(\frac 1 2)$ which is normal with mean $0$ and variance $9(\frac 7 2)+4 \frac 1 2=\frac {67} 2$.

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