Skip to main content
Quantitative Finance

Return to Question

added 87 characters in body
Source Link
sai
sai
  • 21
  • 2

We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs

edit 1: it would be helpful if the explanation is more intuitive than mathematical

We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs

We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs

edit 1: it would be helpful if the explanation is more intuitive than mathematical

Source Link
sai
sai
  • 21
  • 2

standard brownian vs brownian motion

We say Xt with paramters (µ,σ) is brownian process if (Xt-s - X t) ~N (µs,σ2 s) AMONG other conditons .

Here we don't speak about any particular distribution for X t. We only say it is a brownian motion and its increments are normally distributed.

But when it comes to standard brownian motion ( Wt) , why do we say it has a normal distribution i.e Wt ~ N(0,t).

Does that mean I can say any brownian motion process Xt with parameters is µ,σ is also normally distributed N(µt,σ2 s)?

I am new to this topic and if the question does not have a logic, please enlighten with your inputs