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.
1.Suppose that X~B(1,∏). We sample n times and find n1 ones and n2=n-n1zeros
a) What is ML estimator of ∏?
b) What is the ML estimator of ∏ given 1/2≤∏≤1?
c) What is the probability ∏ is greater than 1/2?
d) Find the Bayesian estimator of ∏ under quadratic loss with this prior
2. The attempt...
Homework Equations
L(x,p) = \prod_{i=1}^npdf
l= \sum_{i=1}^nlog(pdf)
Then solve \frac{dl}{dp}=0 for p (parameter we are seeking to estimate)
The Attempt at a Solution
I know how to do this when we are given a pdf, but I'm confused how to do this when we have a sample.
1. Let the pdf of X,Y be f(x,y) = x^2 + \frac{xy}{3}, 0<x<1, 0<y<2
Find P(X+Y<1) two ways:
a) P(X+Y<1) = P(X<1-Y)
b) Let U = X + Y, V=X, and finding the joint distribution of (U,V), then the marginal distribution of U.
The Attempt at a Solution
a) P(X<1-Y) = ?
P(x<1-y) = \int_0^1...
1. Essentially what I'm trying to do is find the asymptotic distributions for
a)
Y2
b) 1/Y and
c) eY where
Y = sample mean of a random iid sample of size n.
E(X) = u; V(X) = σ2
Homework Equations
a) Y^2=Y*Y which converges in probability to u^2,
V(Y*Y)=\sigma^4 + 2\sigma^2u^2
So...
That is the way we are instructed to "name" it in class. f(x) is the joint density function of f(x,y).
fx(x) is equivalent, but 99.9% of the time the Professor uses f(x).
I'm still stuck on whether we should be integrating with respect to y or x (use dy or dx).
Intuitively, dy makes more...
Well, my thinking was that the solution for V(Y|X) is not dependent on the value of y, thus we would only need to use the marginal dist f(x) = \int_{-\infty}^{\infty} f(x,y)dy
Even though V(Y|X) contains no y, should we still use the joint pdf?
Moreover, I started thinking that we should be...
1. Given f(x,y) = 2, 0<x<y<1, show V(Y) = E(V(Y|X)) + V(E(Y|x))
Homework Equations
I've found V(Y|X) = \frac{(1-x)^2}{12} and E(Y|X) = \frac{x+1}{2}
The Attempt at a Solution
So, E(V(Y|X))=E(\frac{(1-x)^2}{12}) = \int_0^y \frac{(1-x)^2}{12}f(x)dx, correct?
Hmm, I thought about that but when I was thinking about it it seemed difficult to calculate E(XY|X=0) for example. That is probably an oversight of mine, though.
But, cov(x,y) = \frac{p(p-1)}{2} is correct?
This is where I am now:
f(y|x=0) = 1/2 for 0<y<2 and f(y|x=1) = 1 0<y<1--> uniform distribution --> E(Y|x=0) = 1 ; E(Y|x=1) = 1/2
Also, E(Y) = E(Y|x=0)P(x=0) + E(Y|x=1)P(x=1) = 1-p/2
So, P(x=0,y) = (1-p)/3 for any y=0,1,2
P(x=1,y) = 1/2 for y=0,1 and 0 for y=2
Then,
cov(x,y) = \sum_{y=0}^2...