Problems in Probability

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Fall 2011 Homework 5: Question 5
Let Z ~ N(0, 1). A measuring device is used to observe Z, but the device can only handle positive values, and gives a reading of 0 if ; this is an example of censored data. So assume that X = Z is observed rather than Z, where is the indicator of Z > 0. Find E(X) and Var(X).
Solution: E(X) = 1/(square root of 2pi). Var(X) = (1/2) - (1/2pi). Note that X is neither purely discrete nor purely continuous (it is a mixed distribution), since X = 0 with probability 1/2 and P(X = x) = 0 for x not equal to 0. So X has neither a PDF nor a PMF; but LOTUS still works, allowing us to work with the PDF of Z to study expected values of functions of Z. Miracle check: The variance is positive, as it should be. It also makes sense that the variance is substantially less than 1 (which is the variance of Z), since we are reducing variability by making the r.v. 0 half the time, and making it nonnegative rather than roaming over the entire real line.
"Mathematics is the logic of certainty, but statistics is the logic of uncertainty."
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