I THINK I may have got it. I basically looked at what we had, and what we need.
In order for us to get back a0, for example, we need:
\frac{1}{2L+1-\frac{I_2^{2}}{I_4}}\sum_{i=-L}^{i=L}a_0(1-i^2\frac{I_2}{I_4}) = a_0
Well, let's multiply through...
I mean, one way to write what we have is
\frac{1}{2L+1-I_2^{2}/I_4} \sum_{i=-L}^{i=L} (1-i^2\frac{I_2}{I_4}) [a_0 + a_1 (t+i) + a_2 (t+i)^2 + a_3 (t+i)^3]
1. Show that applying a second-order weighted moving average to a cubic polynomial will not change anything.
X_t = a_0 + a_1t + a_2t^2 + a_3t^3 is our polynomial
Second-order weighted moving average: \sum_{i=-L}^{i=L} B_iX_{t+i}
where B_i=(1-i^2I_2/I_4)/(2L+1-I_2^{2}/I_4)
where...
Ahh yes, I thought about that also. It's nice to see I wasn't off-base by considering that way.
Why does it not make statistical sense to use the L I suggested (which is based off the fact that X+Y~B(12,p) ?
\frac{dL}{dp} = 8p^7(1-p)^4 - 4p^8(1-p)^3=0
p^7(1-p)^3[8(1-p)]-4p]=0
8-8p-4p=0 ignoring p=0,p=1
8=12p \Rightarrow p=8/12=2/3
Did I miss something?
Note: I screwed up my fraction simplification previously if that's what you meant.
1. Suppose X~B(5,p) and Y~(7,p) independent of X. Sampling once from each population gives x=3,y=5. What is the best (minimum-variance unbiased) estimate of p?
Homework Equations
P(X=x)=\binom{n}{x}p^x(1-p)^{n-x}
The Attempt at a Solution
My idea is that Maximum Likelihood estimators are...
Yes, cov(X,Y) = E(XY)-E(X)E(Y).
Moreover, E(XY) = E(XY|X=0)P(X=0) + E(XY|X=1)P(X=1)
Now, find P(X=0), P(X=1) (this should be easy).
But, you are asking about how to find E(Y|X=1)? Well, we have a formula for f(Y|X=1) don't we? It's a horizontal line at 1 from 0 to 1. Then, the expected value...
Well, based off the graph of \pi^{n_1}(1-\pi)^{n_2} with several different n1 and n2 values plugged in that the best choice would be \pi=n1/n when 1/2≤n1/n≤1, else we choose \pi=1/2 since we usually look at the corner points (1/2 and 1)
What do you mean fail?
Intuitively, \pi_{ML}=\frac{n_1}{n} would "fail" in the case that it is \frac{n_1}{n} < 1/2
But, I'm not sure what our solution must be then if it fails.