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How do you update the probability density function given that the value is greater than a constant?
You want to know the probability that the difference between the random variables is negative. To get PDF of the sum of the two variables, you simply take the convolution of their PDFs. So to get the PDF of their difference, convolve the PDF of one with the PDF of the negation of the other. If you have PDFs f(x) and g(x), the PDF for the difference between the variables is f(x)*g(-x). Just integrate this convolution from 0 to \infty to find the probability that the variable drawn from the distribution represented by f is larger. Example. let f(x) = 2x and g(x) = 1 be two PDFs valid for x between 0 and 1, and 0 elsewhere. For x between 0 and 1, the convolution is (f(x)*g(-x))(t) = (t)=1-t^2. Above 1, itâs 0. Integrating this for all positive values, we get 2/3.
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Alter PDF: All You Need to Know
So the probability of a value between 0 and 0.5 being between 0 and 0.2 is 0.8/3 or 0.0083%. Similarly, with 1 and 1.2, we have 0.5/6, 1/3, 2/9, etc. Now let's assume that this is one extreme case where the difference becomes arbitrarily large (we already know that with any nonzero value, the result is random). So what's the most probable value? Well, given the distribution f and its PDF over x, it's 0.002f(x)=0.0175, so the probability that the value is between 0 and 0.5 is (0.002f(-x)) + 0.0175f(0.5=x). You can check which is better in the calculator below. Now that we've found the limit of the normal distribution, we can look at what the maximum difference is. This is easily done using the mean squared error. The mean squared error is the derivative of the distribution, so f'(x) = f(x-s) where s is the standard deviation. Here, s=30.8. (There.