The use of dt in brownian motion simulations.

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Let's say that I have daily data over three years of some stochastic process. I use Maximum likelihood method to estimate the process. Let's Say its a Ornstein Uhlenbeck Process, now I have often wondered why if i have daily data the increment dt becomes 1/365 $$dx(t)=\sigma dw(t)+\theta dt (\mu -x(t))$$

Now if I have data in daily intervals and I use MLE to find the parameters of this process, or regression I have always wondered why I see examples using dt to become 1/365. I if we have Wiener Process with following conditions I can agree that standard deviation is the square root of the variance.

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Now from this I believe that:

$$\sqrt{\sigma ^2 t}=\sigma \sqrt{t}$$

Now I have often seen that : $$\sigma dw(t)$$

being written as

$$\sigma *\sqrt{\text{dt}}*\text{random}(\text{mean}=0,\text{variance}=1)$$

when simulating.

But regardless of if I am simulating a year or 3 years they always use dt=1/365 but I just dont understand why just because I have daily I need to divide by 365. In other words since I am estimating the parameters over a three year period and I have a daily time period, and I am going to use the parameters to simulate daily prices where is the requirement to divide by 365 coming from.

Of course I understand that 365 is number of days in a year, I just dont see the connection since I am estimating the parameters over daily data in a 3 year period why I would need to define dt as 1/365.

I would expect the yearly volatility to be $\sigma \sqrt{365}$ if $\sigma$ was daily volatility.

I would really appreciate any help I could get including a explanation or pointers to where I have gone wrong in terms of what I have written.

I have spent a long time trying to understand this and I would really appreciate any help I could get. I am very happy to edit the question if it would make it more understandable.