It's not exactly wrong to think of a PDF as relative likelihood, but you can be more precise with language and avoid any ambiguity by thinking of it as what it's called: probability density. It then follows from high school calculus that P(A < X < B) = CDF(B) - CDF(A). The cumulative distribution function (CDF), as a function of x, is the integral of the PDF from -infinity to x, and therefore CDF(x) = P(-infinity < X < x). And the integral limits are important too. It's important to use unambiguous variables here. the probably of drawing a value between A and B. Rather, the probability density function (PDF) is the function that when integrated from A to B gives you P(A < X < B), i.e. For a continuous random variable X, the probability P(X=x) of drawing a particular value x is exactly zero. Plugging some number, x, into the pdf f(x) = P(X=x) which is the relative likelihood of seeing some value x. You've got quite a long wrong here and it's probably causing more confusion for OP. it uses -inf for the lower bound of the integral for each value in the array.Īgain, read up on pdf’s and cdf’s if you do not understand what they are. If x is an array that isn’t an interval, then it returns P(X <= x) for each value in the array. If x is an interval, then you’re finding P(-1 <= X <= 1), which simply changes the bounds of the integral. The function normcdf(x) returns the P(X <= x) so normcdf(0) = 0.5 which is the integral of the pdf from -inf to 0. If x is an array, it returns the relatively likelihood for each of the values in the array (not a difference between them) The function normpdf(x), where x is a single value or array of values, simply returns the relatively likelihood of x, like we said above. In terms of these specific matlab functions: To get the probability of a RV being in a certain domain of x, you need to integrate the pdf.
Let the pdf be some function f(x) where the domain is -inf 1) Ignore mu and sigma, those arguments simply adjust the computation to match a specific normal distribution with those parameters. Okay, it doesn’t look like you received the answer you were looking for so I’m following up.
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