Formula for the expected value of a function

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The expected value of a continuous random variable is defined as E(x) = ∫ x·f(x) dx. For a function g(x) of a random variable X, the expected value is E(Y) = ∫ g(x)·f(x) dx, not E(Y) = ∫ g(x)·f(g(x)) dx. This distinction arises because Y = g(X) is itself a random variable, and the probability density of Y is not simply f(g(x)). The correct formulation emphasizes that the expectation involves the probability density of the original variable X, not the transformed variable. Understanding this concept is crucial for accurately calculating expected values in probability theory.
gsingh2011
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The formula for the expected value of a continuous random variable is E(x) = \int_{-\infty}^{\infty} x\cdot f(x). This leads me to believe that the expected value of a function g(x) is E(x) = \int_{-\infty}^{\infty} g(x)\cdot f(g(x)). However, the correct formula is E(x) = \int_{-\infty}^{\infty} g(x)\cdot f(x). Why is this?
 
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You're talking about a "function of a random variable" (which is itself a random variable) not an ordinary function. The definition of the expectation of a random variable involves an integrand that is the product of the value of the variable times the probability density of the random variable evaluated at that value. If X is a random varaible with density f(x) and Y = g(X) is a function of the random variable X then it is not generally true that the probability density of Y is f(g(x)). Hence f(g(x)) is not the appropriate term to use in the integrand when computing the expectation of Y.
 
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