# Why Cant We Directly Find the PDF of the Transformation of Random?

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## Why can't we directly find the PDF of the transformation of random variables, say g(X) from the random variable X. Why do we have to first convert PDF into CDF and then have to differentiate to get to the PDF of the transformed random variable?

PDF is used to assign the probability of a random variable,falling within a range of values . Its used for a continuous random variable like 1.3,1.4… Its probability is given by taking integral of the variable’s PDF over that range. In mathematical term, The probability density function ("p.d.f.") of a continuous random variable X with support S is an integrable function f(x) satisfying the following. (1) f(x) is positive everywhere in the support S, that is, f(x) > 0, for all x in S (2) The area under the curve f(x) in the support S is 1, that is. ∫Sf(x)dx=1∫Sf(x)dx=1 (3) If f(x) is the p.d.f. of x, then the probability that x belongs to A, where A is some interval, is given by the integral of f(x) over that interval, that is. P(X∈A)=∫Af(x)dx PMF is used to assign the probability of a discrete random variable,which is exactly equal to a number like 1,2,3… In mathematical form, The probability mass function, f(x) = P(X = x), of a discrete random variable X has the following properties. All probabilities are positive. fx(x) ≥ 0. Any event in the distribution (e.g. “scoring between 20 and 30”) has a probability of happening of between 0 and 1 (e.g. 0% and 100%). The sum of all probabilities is 100% (i.e. 1 as a decimal). Σfx(x) = 1. An individual probability is found by adding up the x-values in event A. P(X Ε A) = summation f(x)(xEA) CDF gives the area under PDF upto X values we specify. In mathematical form, Definition. The cumulative distribution function ("c.d.f.") of a continuous random variable X is defined as. F(x)=∫x−∞f(t)dtF(x)=∫−∞xf(t)dt for −∞ < x < ∞.

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