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Unformatted text preview: eometric distribution with parameter p. The fact that
the sum of the probabilities is one,
∞ (1 − p)k−1 p = 1,
k=1 is a consequence of taking x = 1 − p in the formula (1.4) for the sum of a geometric series.
By diﬀerentiating each side of (1.4), setting x = (1 − p), and rearranging, we ﬁnd that E [Y ] =
k=1 kpY (k ) = p . For example, if p = 0.01, E [Y ] = 100.
Another, more elegant way to ﬁnd E [Y ] is to condition on the outcome of the ﬁrst trial. If
the outcome of the ﬁrst trial is one, then Y is one. If the outcome of the ﬁrst trial is zero, then
Y is one plus the number of additional trials until there is a trial with outcome one, which we
call Y . Therefore, E [Y ] = p · 1 + (1 − p)E [1 + Y ] However, Y and Y have the same distribution,
because they are both equal to the number of trials needed until the outcome of a trial is one. So
E [Y ] = E [Y ], and the above equation becomes E [Y ] = 1 + (1 − p)E [Y ], from which it follows that
E [Y ] = p .
The variance of Y can be fou...
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This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.
- Spring '08
- The Land