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Let $X(t), t \geq 0$ be a Brownian motion process with drift parameter $\mu$ and variance parameter $\sigma^{2}$ for which $X(0) = 0$. Show that $-X(t), t \geq 0$ is a Brownian motion process with drift parameter -$\mu$ and parameter $\sigma^{2}$

I'm not too sure about how to approach this question. I think that we need to show that $-X(t)$ satisfies the definition of a Brownian motion, namely that $X(0)$ is a given constant, and for all positive $y$ and $t$, $X(t + y) - X(y)$ is independent of the process up to time $y$ and has a normal distribution with parameters $t\mu$ and $t\sigma^{2}$.

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  • $\begingroup$ Which one of the properties you have difficulty with ? $\endgroup$ – Kavi Rama Murthy Mar 8 at 7:20
  • $\begingroup$ For the first property, I think I can just say $-X(0)= -0 = 0$, so this property is satisfied. Is this correct? I'm struggling with the second property. $\endgroup$ – user651921 Mar 8 at 7:23
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For the first point, you are indeed right, since Brownian motion with drift starts at $0$ with probability $1$.

As for the other properties, note that if we let $Y(t)=-X(t)$, then you have that

$$Y(t) - Y(s) = X(s)-X(t) = -(X(t)-X(s)) \quad \text{for} \quad t>s$$

Using the properties of Brownian motion (distribution, independence, stationarity of increments) and the fact that the transformation applied is linear, can you see what the distribution of this last quantity is?

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  • $\begingroup$ So, how can I show that the distribution is normal with drift parameter $-\mu$ and $\sigma^{2}$? $\endgroup$ – user651921 Mar 8 at 12:16
  • $\begingroup$ You do know the distribution of $X(t) -X(s)$, because of the definition of Brownian motion with drift. What happens when you apply a linear transformation to this distribution? $\endgroup$ – Easymode44 Mar 8 at 12:19
  • $\begingroup$ I am not familiar with linear transformations. I know that $X(t) - X(s)$ follows normal distribution with mean $(t - s)\mu$ and variance $(t - s)\sigma^{2}$ $\endgroup$ – user651921 Mar 8 at 12:22
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    $\begingroup$ In other words: if $Z$ is normal with mean $a$ and variance $b$ ($Z\sim\mathcal{N}(a,b)$) what is the distribution of $-Z$? $\endgroup$ – Minus One-Twelfth Mar 8 at 12:25
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    $\begingroup$ Since you already know that $Z:= X(t) - X(s)$ is normal with mean $a:= (t-s)\mu$ and variance $(t-s)\sigma^2$, you can apply that result about $-Z$. Or do you mean you think you have to show that result about $-Z$? I don't think you would be expected to show that here -- it is a very basic fact about normal distributions and would be assumed knowledge at this point since you are dealing with more complicated things now (like Brownian motion). $\endgroup$ – Minus One-Twelfth Mar 8 at 12:27

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