Unformatted text preview: d Y be the number of
successes in the last m. By (2.13), X and Y independent. Clearly their sum is
Formula (A.16) generalizes in the usual way to continuous distributions:
regard the probabilities as density functions and replace the sum by an integral.
fX +Y (z ) = fX (x)fY (z x) dx
Example A.17. Let U and V be independent and uniform on (0, 1). Compute
the density function for U + V .
Solution. If U + V = x with 0 x 1, then we must have U x so that
0. Recalling that we must also have U 0
fU +V (x) =
1 · 1 du = x when 0 x 1
0 If U + V = x with 1 x 2, then we must have U
x 1 so that V 1.
Recalling that we must also have U 1,
fU +V (x) =
1 · 1 du = 2 x when 1 x 2
x1 Combining the two formulas we see that the density function for the sum is
triangular. It starts at 0 at 0, increases linearly with rate 1 until it reaches the
value of 1 at x = 1, then it decreases linearly back to 0 at x = 2. A.3 Expected Value, Moments If X has a discrete distribution, then the expected value of h(X ) is
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.
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