/sci/ - Science & Math

[Anonymous No.4615703]

So this is hugely annoying me after my econ homework.

Let d_t = a * d_(t-1) + v_t; 0 < a < 1, v_t is white noise with mean 0 and variance sigma^2

Wikipedia says that since |a| < 1, E(d_t) = u for all t, and thus:

E(d_t) = a * E(d_(t-1)) + E(v_t)
u = a*u + 0
u = 0

However, if we proceed recursively:

d_1 = a * d_0 + v_1
d_2 = a * (a * d_0 + v_1) + v_2
...
d_n = a^n * d_0 + a^(n-1) * v_1 + a^(n-2) * v_2 + ... + a * v_(n-1) + v_n

E(d_n) = E(a^n * d_0) since E(v_t) = 0
E(d_n) = a^n*d_0

Which is evidently not 0.

The fuck is going on?

[Anonymous No.4615723]

This is when LaTeX may come in handy

[Anonymous No.4615739]

What the hell is E?

[Anonymous No.4615741]

>>4615739

expected value

[Anonymous No.4615751]

>>4615741
Over all time?

Then yeah, if a<1 then you'll head for an expected value of zero (exponential decay with added noise). Lemme look through your second section again now.

[Anonymous No.4615758]

>E(d_n) = a^n*d_0
>which is evidently not zero
In the limit of n going to positive infinity, yes, this is zero. a<1, remember?

[Anonymous No.4615762]

>>4615751

But the confusion I have here is that that trend contradicts the reasoning used in the first place.

If E(d_t) = u and E(d_(t+1)) < u (as is evident from the recursive method), then E(d_(t-1)) != u.