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# The random 1 999 variable x is equal to 100 with

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Unformatted text preview: he function g (u) = u2 + 3u. Using LOTUS yields E [X 2 + 3 X ] = (u2 + 3ui )pX (ui ) i i u2 pX (ui ) i = i + 3ui pX (ui ) i = E [X 2 ] + 3E [X ]. This example illustrates that expectation is a linear operation. Example 2.2.6 Suppose two fair dice are rolled. Find the pmf and the mean of the product of the two numbers showing. Solution Let X1 denote the number showing on the ﬁrst die and X2 denote the number showing 1 on the second die. Then P {(X1 , X2 ) = (i, j )} = 36 for 1 ≤ i ≤ 6 and 1 ≤ j ≤ 6. Let Y = X1 X2 . The easiest way to calculate the pmf of Y is to go through all 36 possible outcomes, and add the probabilities for outcomes giving the same value of Y. A table of Y as a function of X1 and X2 is shown in Table 2.1. There is only one outcome giving Y = 1, so pY (1) = 1/36. There are two outcomes giving Y = 2, so pY (2) = 2/36, and so on. The pmf of Y is shown in Figure 2.3. One way to compute E [Y ] would be to use the deﬁnition of expectation and the pmf of Y shown. An alternative that we take is to use LOTUS. We have that Y = g (X1 , X2 ), where g (i,...
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