## Wiener Process Properties

I have a simple question about the intuition behind property 1 of a Wiener Process. It says in my textbook that the change in a variable z that follows a Wiener Process is:

$$δz=ε\sqrt{δt}$$

where ε is a random drawing from a $$\Phi(0,1)$$

Now I think $$\sqrt{δt}$$ is supposed to be the standard deviation of a random variable which follows a normal distribution with a standard deviation of 1 during one year.

My question now is, if δ^1/2 is the standard deviation of a normally distributed random variable, why is the random drawing from another normal distribution necessary or basically why do I have to multiply ε with δt^1/2?
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 Quote by Polymath89 I have a simple question about the intuition behind property 1 of a Wiener Process. It says in my textbook that the change in a variable z that follows a Wiener Process is: $$δz=ε\sqrt{δt}$$ where ε is a random drawing from a $$\Phi(0,1)$$ Now I think $$\sqrt{δt}$$ is supposed to be the standard deviation of a random variable which follows a normal distribution with a standard deviation of 1 during one year. My question now is, if δ^1/2 is the standard deviation of a normally distributed random variable, why is the random drawing from another normal distribution necessary or basically why do I have to multiply ε with δt^1/2?
If you don't multiply your delta z will be too big. Suppose t is really small then certainly delta z does not have distribution N(0,1).
 Thanks for your answer. You're right in saying that δz would be really big, if the time change is small, but that's not a really satisfying answer for me. I want to know why it's normally distributed, not why it's not normally distributed if you have a different form.

## Wiener Process Properties

 Quote by Polymath89 Thanks for your answer. You're right in saying that δz would be really big, if the time change is small, but that's not a really satisfying answer for me. I want to know why it's normally distributed, not why it's not normally distributed if you have a different form.
why what is normally distributed ?

 Quote by Polymath89 I have a simple question about the intuition behind property 1 of a Wiener Process. It says in my textbook that the change in a variable z that follows a Wiener Process is: $$δz=ε\sqrt{δt}$$ where ε is a random drawing from a $$\Phi(0,1)$$ Now I think $$\sqrt{δt}$$ is supposed to be the standard deviation of a random variable which follows a normal distribution with a standard deviation of 1 during one year. My question now is, if δ^1/2 is the standard deviation of a normally distributed random variable, why is the random drawing from another normal distribution necessary or basically why do I have to multiply ε with δt^1/2?
If you take Wiener process and break it up into any amount of intervals eg 2,3,1000,1000000 etc. Then withen each interval you have in effect another Wiener process. In this sense it is like a fractal. No matter how small the interval, you always have another Wiener process. So one thing you can ask your self is how would you simulate Wiener process on a computer.

Recognitions:
From the this point of view the intutiive idea of a Wiener process is that it is a phenomenon whose increments can be analyzed as independent normal random variables at increasingly small time intervals. Of course, if the variance of the process measured at t = 10, 20, 30,... is $\sigma_{10}$, you can't say the variance of the process at t = 1, 2, 3,... is one tenth of that variance. To have the variance at 10 small increments produce the correct variance at one large increment, you must consider how the variances of a sum of independent random variables add. That's essentially where the square root comes in.