Then p x1 x2 i j 36 for 1 i 6 and 1 j 6

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Unformatted text preview: t of nonzero values of pY is as follows: u pY (u) 0 1/8 1 2/8 4 2/8 9 1/8 16 1/8 25 1/8. Therefore, E [Y ] = 0·1+1·2+4·2+9·1+16·1+25·1 8 = 60 8 = 7.5. You may have noticed, without even thinking about it, that there is another way to compute E [Y ] in Example 2.2.4. We can find E [Y ] without finding the pmf of Y. Instead of first adding together some values of pX to find pY and then summing over the possible values of Y, we can just directly sum over the possible values of X, yielding 1 E [Y ] = {(−2)2 + (−1)2 + 02 + 12 + 22 + 32 + 42 + 52 }. 8 The general formula for the mean of a function, g (X ), of X , is E [g (X )] = g (ui )pX (ui ). i This formula is so natural it is called the law of the unconscious statistician (LOTUS). 2.2. THE MEAN AND VARIANCE OF A RANDOM VARIABLE 25 Table 2.1: Table of Y vs. X1 and X2 for Y = X1 X2 . X1 \X2 1 2 3 4 5 6 123 123 246 369 4 8 12 5 10 15 6 12 18 4 4 8 12 16 20 24 5 6 5 6 10 12 15 18 20 24 25 30 30 36. Example 2.2.5 Let X be a random variable with a pmf pX and consider the new random variable Y = X 2 + 3X. Use LOTUS to express E [Y ] in terms of E [X ] and E [X 2 ]. Solution: So Y = g (X ) for t...
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