Think of this from the perspective of an observer

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 . p2 1 p, 2−p . Therefore, Var(Y ) = p2 √ 1−p Var(Y ) = p . 1 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...
View Full Document

Ask a homework question - tutors are online