Dan's Segment 24

From Computational Statistics (CSE383M and CS395T)
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To Calculate

1. We know that the sum of N normal deviates is normally distributed itself, so we can easily calculate the mean and variance of X.

<math> <X> = \sum_{k=1}^{20} \alpha_k <T_k> = 0 </math>

And

<math> Var(x) = \sum_{k=1}^{20} \alpha_k^2 Var(T_k) = \sum_{k=1}^{20} \alpha_k^2 </math>

So we can easily see that X is <math>N(0,\sum_{k=1}^{20} \alpha_k^2) </math>distributed.

This means the t-value is <math> t = \frac{X}{\sqrt{\sum_{k=1}^{20} \alpha_k^2}} </math> which follows the N(0,1) distributions t-value. Thus, we can see that <math>t^2</math> is chisquare(1) distributed.


To Think About

1. This was the class activity, and it was completed with Lori and I think Silu.

2. In general for the Cauchy distribution the mean should not follow the "universal rule" because its first moment doesn't exist. In contrast, since the median doesn't depend on the first moment and is resistant to outliers, it should remain constant with the number of trials and thus the N^(1/2) rule should hold.