10 for 75 r r r p expectation for a continuous

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% 10 for 75 . 0 ) ( r r r P
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Expectation For a continuous probability function The idea is the same as for a discrete PDF. We just integrate across all possible values rather than sum over the discrete values. - = dr r rP r E ) ( ] [
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Expectation At times we are interested in the expectation of a function of a random variable. Example E[r]=.65*(.08)+(.35)*(-.10)=0.017 What is E[r 2 ]? E[r 2 ]=.65*(.08 2 )+.35*(-.10 2 )=0.008 What is E[3r+5]? E[3r+5]=.65*(3*.08+5)+.35*(3*(-.10)+5)=5.051 - = = = % 10 for 35 . 0 % 8 for 65 . 0 ) ( r r r P
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Expectation Given a discrete PDF we now know how to calculate the expected value of a random variable, and the expected value of a function of a random variable. PDFs are like number generating machines Sometimes, we don’t know exactly everything about the machine, but we observe the output of the machine. Output = numbers that the machine generates By looking at the output the machine generates, we can get some idea about what kind of machine (PDF) is generating the numbers.
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Expectation Example E[r]=.65*(.08)+(.35)*(-.10)=0.017 Suppose the above PDF generates 100 numbers, and we take the sample average. 100 is much less than infinity: the average of these 100 numbers is not likely to be exactly 0.017, however, it should be close. The difference arises because of sampling error - the fact that our sample does not contain an infinite number of observations. - = = = % 10 for 35 . 0 % 8 for 65 . 0 ) ( r r r P
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Expectation Given a sample of independent outcomes from the PDF, the sample average is statistically the “best” estimate of the true expectation given the information you have. Side note: by “best” we mean Unbiased Consistent Efficient These terms are part of the CFA curriculum and we may discuss them more in depth later.
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Estimation Suppose we observe the following numbers generated by a PDF: (.10, .10, .10, .10, .10,.10, -.08,-.08,-.08,-.08). Estimate E[r] Ê[r]=(.10+.10+ .10+.10+.10+.10-.08-.08-.08-.08)/10=0.028 Note that it looks like these numbers may have been generated from a PDF with two possible outcomes such that Also note that .65*(.10)+.35*(-.08)=0.037 - = = = % 8 for 35 . 0 % 10 for 65 . 0 ) ( r r r P
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Estimation Suppose we observe the following numbers generated by a PDF: (.10, .10, .10, .10, .10,.10, -.08,-.08,-.08,-.08). Estimate E[r 2 ] Ê[r 2 ]= [.10 2 +.10 2 + .10 2 +.10 2 +.10 2 +.10 2 +(-.08) 2 +(-.08) 2 +(-.08) 2 +(-.08) 2 ]/10=0.009 Also note that .65*(.10 2 )+.35*(-.08) 2 =0.00874
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Expectation Summary Given a discrete PDF we can find the true expectation by We can also find the true expectation of a function of a random variable, g(r) as
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  • Fall '10
  • BrianBoyer
  • Variance, Probability theory, probability density function, Cumulative distribution function

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