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Limiter: Find an expression for the mean and variance of yₑ = g(x) for an arbitrary continuous random variable x.

a) Mean: E(g(x)), Variance: Var(g(x))
b) Mean: μ, Variance: σ²
c) Mean: g(μ), Variance: g(σ²)
d) Mean: E(x), Variance: Var(x)

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User Ttimasdf
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1 Answer

4 votes

Final answer:

The correct expressions for the mean and variance of yₑ = g(x), where x is a continuous random variable, are E(g(x)) for the mean, and Var(g(x)) for the variance: option (a).

Step-by-step explanation:

Expression for the Mean and Variance of yₑ = g(x)

To find an expression for the mean and variance of a function of a continuous random variable, yₑ = g(x), you must consider the properties of expected value and variance operations. The mean of yₑ is the expected value of g(x), denoted as E(g(x)), which is the long-term average you would expect if you could repeat an experiment an infinite number of times. The variance of yₑ is denoted by Var(g(x)), which measures the spread of g(x) around its mean.

The correct expression for the mean of yₑ = g(x) is not g(μ), nor is it E(x). It is E(g(x)), and this is because the mean of a transformed random variable is the expected value of that transformation applied to the original variable, not just the transformation applied to the mean of the original variable. Similarly, the variance of yₑ = g(x) is Var(g(x)), and not g(σ²) nor Var(x), because the variance is the expected value of the squared deviation of g(x) from its mean, which takes into account the nonlinearity of the transformation g.

answered
User Josh Buhler
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8.4k points
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