Recent content by Scootertaj

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    Weighted Moving Average of Cubic

    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...
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    Weighted Moving Average of Cubic

    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]
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    Weighted Moving Average of Cubic

    I'm unsure as to how you rearranged the weights to get B_i = A - B i^2, would you mind clarifying what A and Bi2 are?
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    Weighted Moving Average of Cubic

    Sorry, it just follows the same pattern as I2: I_4=\sum_{i=-L}^{i=L} i^{4}
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    Weighted Moving Average of Cubic

    L is any arbitrary number. For any L, this should be true.
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    Weighted Moving Average of Cubic

    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...
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    Estimate p from sample of two Binomially Distributions

    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) ?
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    Estimate p from sample of two Binomially Distributions

    \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.
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    Estimate p from sample of two Binomially Distributions

    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...
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    Covariance - Bernoulli Distribution

    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...
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    Covariance - Bernoulli Distribution

    Yes, you should get an answer in terms of p. Just use the E(XY) formula above and recall the formula for cov(X,Y).
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    Covariance - Bernoulli Distribution

    E(XY)=E(xy|x=0)P(x=0) + E(xy|x=1)P(x=1)
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    Maximum Likelihood Estimator + Prior

    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)
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    Maximum Likelihood Estimator + Prior

    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.
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