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Unformatted text preview: nd similarly. We could diﬀerentiate each side of (1.4) twice, set
x = 1 − p, and rearrange. Here we take the alternative approach, just described for ﬁnding the
mean. By the same reasoning as before,
E [Y 2 ] = p + (1 − p)E [(1 + Y )2 ] or E [Y 2 ] = p + (1 − p)E [(1 + Y )2 ].
Expanding out, using the linearity of expectation, yields
E [Y 2 ] = p + (1 − p)(1 + 2E [Y ] + E [Y 2 ]).
Solving for E [Y 2 ], using the fact E [Y ] =
E [Y ]2 = 1−p
. Therefore, Var(Y ) =
Var(Y ) = p .
the mean E [Y ] = p is very large,
√ yields E [Y 2 ] = The standard deviation of Y is σY = It is worth remembering that if p is very small,
standard deviation is nearly as large as the mean (just smaller by the factor E [Y 2 ] −
and the 1 − p). Example 2.5.1 Suppose a fair die is repeatedly rolled until each of the numbers one through six
shows at least once. What is the mean number of rolls?
Solution. The total number of rolls, R, can be expressed as R = R1 + . . . + R6 , where for 1 ≤ i ≤ 6,
Ri is the number of rolls made after i − 1 distinct nu...
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- Spring '08
- The Land