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3.2 The Law of the Unconscious Statistician

Recall that for a continuous rv, Y,

𝖤[Y]=-ufY(u)du.

If Y=g(X) then how do we know that this is equivalent to

𝖤[Y]=-g(u)fX(u)du?

A similar result is proved for discrete rvs in the notes. For simplicity of exposition, we assume here that the continuous rvs X and Y are non-negative.

We first turn the expectation of Y into a double integral. Reversing the order of integration provides a term 𝖯(Y>t), which is equivalent to 𝖯(g(X)>t); returning to the original order of integration produces the required result. Along the way we will prove that (if Y0), 𝖤[Y]=0SY(t)dt, where SY(y) is the survivor function of Y; this is of interest in its own right.

Unnumbered Figure: First link, Second Link

Using the left figure for the first change of order of integration and the right figure for the second change of order, and recalling that y=0y1dt, we have:

𝖤[Y] =0fY(u)udu
=y=0t=0yfY(u)dtdu
=t=0y=tfY(u)dudt
=t=0𝖯(Y>t)dt
=t=0𝖯(g(X)>t)dt
=t=0x:g(x)>tfX(x)dxdt
=x=0t=0g(x)fX(x)dtdx
=x=0fX(x)g(x)dx,

as required.