- #1
Rasalhague
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- 2
Hoel: An Introduction to Mathematical Statistics introduces the following formulas for expectation, where the density is zero outside of the interval [a,b].
[tex]E\left [ X \right ] = \int_{a}^{b} x f(x) \; dx[/tex]
[tex]E\left [ g(X) \right ] = \int_{a}^{b} g(x) f(x) \; dx[/tex]
He says, "Let the random variable g(X) be denoted by Y. Then knowing the density f(x) of X it is theoretically possible to to find the density h(x) of Y. The expected value of g(X) is the same as the expected value of Y; therefore if h(y) is available, the latter expected value can be expressed in the form
[tex]E\left [ Y \right ] = \int_{-\infty}^{\infty} y h(y) \; dy.[/tex]
"By using the change of variable techniques of calculus, it can be shown that this value is the same as the value given by (22) [the 2nd formula I've quoted in this post]."
I've been trying to do this. Let I denote the identity function on [itex]\mathbb{R}[/itex]. Let [itex]f_X[/itex] denote the pdf of the distribution induced by a random variable X, and [itex]F_X[/itex] its cdf. I'm guessing that when the expected value of a distribution is expressed like this in terms of a random variable, [itex]E[X][/itex] is to be understood as [itex]E[P_X][/itex], and [itex]E[g(X)][/itex] as [itex]E[P_{g \circ X}][/itex], where [itex]P_X[/itex] means the distribution induced by the random variable X, given some sample space implicit in the context.
Then expectation is defined by
[tex]E[P_X]=\int_a^b I \cdot f_X,[/tex]
and we must show that
[tex]\int_a^b I \cdot f_{g \circ X} = \int_a^b g \cdot f_X,[/tex]
or do the limits need to be changed? Using the chain rule (integration by substitution) and the identity
[tex]F_{g \circ X}=F_X \circ g,[/tex]
leads me to
[tex]\int_a^b I \cdot f_{g \circ X} = \int_{g(a)}^{g(b)} I \cdot f_X[/tex]
which looks tantalisingly close, but am I going in the right direction?
[tex]E\left [ X \right ] = \int_{a}^{b} x f(x) \; dx[/tex]
[tex]E\left [ g(X) \right ] = \int_{a}^{b} g(x) f(x) \; dx[/tex]
He says, "Let the random variable g(X) be denoted by Y. Then knowing the density f(x) of X it is theoretically possible to to find the density h(x) of Y. The expected value of g(X) is the same as the expected value of Y; therefore if h(y) is available, the latter expected value can be expressed in the form
[tex]E\left [ Y \right ] = \int_{-\infty}^{\infty} y h(y) \; dy.[/tex]
"By using the change of variable techniques of calculus, it can be shown that this value is the same as the value given by (22) [the 2nd formula I've quoted in this post]."
I've been trying to do this. Let I denote the identity function on [itex]\mathbb{R}[/itex]. Let [itex]f_X[/itex] denote the pdf of the distribution induced by a random variable X, and [itex]F_X[/itex] its cdf. I'm guessing that when the expected value of a distribution is expressed like this in terms of a random variable, [itex]E[X][/itex] is to be understood as [itex]E[P_X][/itex], and [itex]E[g(X)][/itex] as [itex]E[P_{g \circ X}][/itex], where [itex]P_X[/itex] means the distribution induced by the random variable X, given some sample space implicit in the context.
Then expectation is defined by
[tex]E[P_X]=\int_a^b I \cdot f_X,[/tex]
and we must show that
[tex]\int_a^b I \cdot f_{g \circ X} = \int_a^b g \cdot f_X,[/tex]
or do the limits need to be changed? Using the chain rule (integration by substitution) and the identity
[tex]F_{g \circ X}=F_X \circ g,[/tex]
leads me to
[tex]\int_a^b I \cdot f_{g \circ X} = \int_{g(a)}^{g(b)} I \cdot f_X[/tex]
which looks tantalisingly close, but am I going in the right direction?