Formula for the expected value of a function

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SUMMARY

The expected value of a continuous random variable is defined by the formula E(x) = ∫_{-∞}^{∞} x·f(x). However, when considering the expected value of a function g(x) of a random variable X, the correct formula is E(g(X)) = ∫_{-∞}^{∞} g(x)·f(x) rather than E(g(X)) = ∫_{-∞}^{∞} g(x)·f(g(x)). This distinction arises because g(X) is itself a random variable, and the probability density of Y = g(X) cannot be expressed as f(g(x)). Understanding this concept is crucial for accurately calculating expected values in probability theory.

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  • Understanding of continuous random variables
  • Familiarity with probability density functions (PDFs)
  • Knowledge of integration techniques in calculus
  • Concept of functions of random variables
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  • Study the properties of probability density functions
  • Learn about transformations of random variables
  • Explore the concept of moment-generating functions
  • Investigate applications of expected value in statistics
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gsingh2011
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The formula for the expected value of a continuous random variable is [itex]E(x) = \int_{-\infty}^{\infty} x\cdot f(x)[/itex]. This leads me to believe that the expected value of a function g(x) is [itex]E(x) = \int_{-\infty}^{\infty} g(x)\cdot f(g(x))[/itex]. However, the correct formula is [itex]E(x) = \int_{-\infty}^{\infty} g(x)\cdot f(x)[/itex]. 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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