Unformatted text preview: 36226 Summer 2010 Homework 1 Due June 30 1. Math review.
(a) Simplify the following expression as much as possible. The xi are positive integers. There
should be no
signs in the result but there will be
signs.
n ln
i=1 ea bxi
xi ! . (1) (b) Find the value of x which maximizes the following function. Be sure to prove that it is
a maximum.
f (x  n, a) = a ln x + (n − a) ln(1 − x)
¯
(c) Let X = 1
n n
i=1 Xi . 0<a<n (2) Show that
n n ¯
¯
(Xi − X )2 + n(X − µ)2 . (Xi − µ)2 =
i=1 (3) i=1 Hint: Start with the left side. The ﬁrst step is a trick. The rest is easy. Do not work
from both sides. Proofs should be linear and sensical.
2. Probability review.
Let X1 and X2 be independent, each with pdf
f (x) = 1
exp{−x/β }I[0,∞) (x)
β β > 0. (4) (a) Find E [X1 ] and V ar[X1 ].
(b) Write down the joint density and integrate to ﬁnd E [3X1 − 2X2 + 10]. Feel free to cite
part (i) if necessary.
(c) Use properties of expectation and the answer to (i) to verify your answer in (ii).
(d) Find V ar[3X1 − 2X2 + 10] in any manner you wish.
3. The art of statistics.
In class, I determined which realization of ones and zeros was random by looking for long
strings of successive ones (or zeros). Let’s formalize this idea. Write down a probability model
for the random string of heads and tails. To do this, deﬁne some random variable(s) and
parameter(s) and articulate a distribution for the random variable(s). Also carefully deﬁne
what it means to see a “long string”. What is complicated about trying to determine the
probability of observing at least one “long string”? This question is intentionally open ended
and should stimulate thinking about the problem. Do NOT try to actually calculate the
probability of “long strings”. 1 ...
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 Summer '09

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