Hi guys, I have one main question with a given example and a couple other ones related to understanding some concepts.

First, here’s the example function:

```
function brownian_paths(; paths=10^3, n=10^5, t=1, S0=100, μ=(S,k) -> 0.0S, σ=(S,k) -> 0.3S)
Δ = t / n
S = randn((n, paths))
S[1,:] .= S0
if n > 1
for i = 2:n
@. S[i,:] = S[i-1,:] + μ(S[i-1,:], (i-1)*Δ)*Δ + σ(S[i-1,:], (i-1)*Δ)*sqrt(Δ)*S[i,:]
end
end
return S
end;
```

I know that I am allocating memory when I create the S array. But as far as I understand from Julia’s website, running the loop starting “@.” makes that whole line do operations in-place on the i-th row of S instead of having to create some new Arrays every time and then assign them to that row.

But instead I see that loop allocates a lot more memory. You can see how much by running the following instead:

```
using TimerOutputs
function brownian_paths(; paths=10^3, n=10^5, t=1, S0=100, μ=(S,k) -> 0.0S, σ=(S,k) -> 0.3S)
reset_timer!()
Δ = t / n
@timeit "Initialize S" S = randn((n, paths))
S[1,:] .= S0
@timeit "Loop" begin
if n > 1
for i = 2:n
@. S[i,:] = S[i-1,:] + μ(S[i-1,:], (i-1)*Δ)*Δ + σ(S[i-1,:], (i-1)*Δ)*sqrt(Δ)*S[i,:]
end
end
end
print_timer()
return S
end;
```

Does anyone have any insight on why this is happening? Might be obvious but idk, been digging all day and couldn’t find an answer to this anywhere.

(If you have any roasts to this function go ahead New to Julia and my first project is to re-code all my option pricing algorithms in it)