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?

CDF stands for cumulative distribution function. It takes as input any real number, and returns as output a number from 0 up to 1. It is defined as \text{cdf}_X(a) = F_X(a) = \mathbb{P}(X \leq a). PDF stands for probability density function. It is a bit trickier to define. When X is a continuous random variable, then \text{pdf}_X(a) = f_X(x) = [\cdf_X(a)]' = d\text{cdf}_X(a)/da. When X is a discrete random variable, then \text{pdf}_X(a) = \mathbb(P)(X = a). So a CDF is a function whose output is a probability. The PDF is a function whose output is a nonnegative number. The PDF itself is not a probability (unlike the CDF), but it can be used to calculate probabilities. When X is a continuous random variable, then we use integration to calculate probabilities using the PDF. \mathbb{P}(X \in A) = \int_{a \in A} f_X(a) \ dx. When X is a discrete random variable, then we use summation to calculate probabilities using the PDF. \mathbb{P}(X \in A) = \sum_{a \in A} f_X(a).

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Convert PDF: All You Need to Know

Begin{document} \label{simple.pdf} \math bf{CDF} \math bf{CDF}(x) = \int_a2f{(X_a)}. \end{document} \label{simple.pdf} \math bf{PDF} \math bf{PDF}(x) = \int_a2f({\Cdf_x}). \end{document} \begin{math par{1}\begin{array}{ll}x \LEQ 1 &\text{if} x > 1 \\ 1 \LEQ 2 &\text{if} x

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