andrew_gelman_stats andrew_gelman_stats-2010 andrew_gelman_stats-2010-341 knowledge-graph by maker-knowledge-mining
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Introduction: I had the following email exchange with a reader of Bayesian Data Analysis. My correspondent wrote: Exercise 1(b) involves evaluating the normal pdf at a single point. But p(Y=y|mu,sigma) = 0 (and is not simply N(y|mu,sigma)), since the normal distribution is continuous. So it seems that part (b) of the exercise is inappropriate. The solution does actually evaluate the probability as the value of the pdf at the single point, which is wrong. The probabilities should all be 0, so the answer to (b) is undefined. I replied: The pdf is the probability density function, which for a continuous distribution is defined as the derivative of the cumulative density function. The notation in BDA is rigorous but we do not spell out all the details, so I can see how confusion is possible. My correspondent: I agree that the pdf is the derivative of the cdf. But to compute P(a .lt. Y .lt. b) for a continuous distribution (with support in the real line) requires integrating over t
sentIndex sentText sentNum sentScore
1 My correspondent wrote: Exercise 1(b) involves evaluating the normal pdf at a single point. [sent-2, score-0.876]
2 But p(Y=y|mu,sigma) = 0 (and is not simply N(y|mu,sigma)), since the normal distribution is continuous. [sent-3, score-0.432]
3 The solution does actually evaluate the probability as the value of the pdf at the single point, which is wrong. [sent-5, score-1.08]
4 I replied: The pdf is the probability density function, which for a continuous distribution is defined as the derivative of the cumulative density function. [sent-7, score-1.276]
5 My correspondent: I agree that the pdf is the derivative of the cdf. [sent-9, score-0.471]
6 b) for a continuous distribution (with support in the real line) requires integrating over the pdf on the interval [a,b]. [sent-14, score-0.69]
7 ] So the probability of drawing exactly y from a normal distribution is 0. [sent-27, score-0.647]
8 The probability of drawing some value around a neighborhood of y, however, is nonzero. [sent-28, score-0.499]
9 Since in this exercise P(Y=y) appears in the denominator of the conditional probability, that probability is undefined. [sent-29, score-0.574]
10 The mistake made in the solution is to substitutes N(1|mu, sigma2) for P(Y=1|mu,sigma2), which is incorrect. [sent-30, score-0.463]
11 (I had to state the probability that an individual’s height was, say, 5’0″, with height a normal r. [sent-35, score-0.634]
12 I ended up “faking” the answer by evaluating the pdf as done in the solution to the exercise here, even though the probability should have been 0. [sent-37, score-1.221]
13 In this case, theta is a discrete random variable and y is continuous. [sent-39, score-0.342]
14 ) If you really want, you could compute the probability that theta=1 given that |y=1| . [sent-43, score-0.358]
15 epsilon, and then take the limit as epsilon approaches 0. [sent-45, score-0.333]
16 I wanted only to find the solution to your problem and was stopped in my tracks by the fact that P(y=1|theta=1) is 0. [sent-47, score-0.33]
17 Indeed, I still don’t know how to solve the exercise on this account — short of computing the limit, as you suggested. [sent-48, score-0.316]
18 For this reason I think the solution as provided is misleading at best, or wrong at worst. [sent-49, score-0.264]
19 Me: If you look at the solution, I write p(y=1) etc, not Pr(y=1), so I am in fact referring to probability densities rather than discrete probabilities. [sent-50, score-0.464]
20 (Short of “knowing” that the true solution can be obtained by making this non sequitur substitution. [sent-54, score-0.326]
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