# Thread: Probability of exact value of normally distributed variable

1. ## nevermind

(disregard the nevermind, I still haven't quite figured it out)

In the solutions manual for Bayesian Data Analysis on problem 1.1b, the author appears to calculate the probability of an exact value for a normally distributed variable by simply plugging it in to the equation.

It's my understanding that this should always be 0, but I haven't done any stats for a few years so I'm assuming I have forgotten something, however I haven't found anything online to the contrary.

http://www.stat.columbia.edu/~gelman...solutions2.pdf

The final form of the equation:
(0.5 * N(1 | 1,2^2)) / 0.5(N(1 | 1,2^2) + N(1 | 2,2^2))

which works out to ~.53.

2. ## Re: Probability of exact value of normally distributed variable

Which piece do you believe should be zero? The final answer or one of the intermediate values?

3. ## Re: Probability of exact value of normally distributed variable

I was thinking that the evaluation for each term with a normal distribution should be 0, which would make the whole thing 0 or indeterminate. It simplifies to 1/(1+e^(-(y-1)/8)), but isn't it still a continuous distribution, which would mean the probability of any exact y is 0?

4. ## Re: Probability of exact value of normally distributed variable

This may or may not help: but as an example if test scores are normally distributed between 0-100. You could calculate the probability of scoring 53 (an exact value) by changing the exact value into an interval 52.5 to 53.5. What's happening is that you're approximating test scores by a normal distribution, but it's clearly a non-zero probability to score exact values (0, 1, 2, ..., 98, 99, 100).

So in your problem, maybe they're turning the p(1) into an interval from 0.5 to 1.5, or maybe from 1 to 1.5.

5. ## The Following User Says Thank You to Mean Joe For This Useful Post:

StrongPrior (06-07-2017)

6. ## Re: Probability of exact value of normally distributed variable

I had the same thought, but there's no mention of something like that and it seems a little out of place. Perhaps I'm being too pedantic.

7. ## Re: Probability of exact value of normally distributed variable

They're using the density value. The argument for why you can do that is basically analagous to Mean Joe's argument as you let the interval get smaller and smaller.

8. ## Re: Probability of exact value of normally distributed variable

Ok, thanks.

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