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Suppose I have the probability space $(X,\mathcal{E},P)$ and a random variable $u : X \to \mathbb{R}$. $u$ induces a probability $Q (B) = P (u^{-1}(B))~\forall B \in \mathcal{B}$, the Borel $\sigma$-algebra on $\mathbb{R}$ and we can consider a new probability space $(\mathbb{R},\mathcal{B},Q)$.

Now, I am having trouble connecting this theory to practical applications. For example, can I let $u$ to be a Standard Normal Random Variable? If so, the expectation of $u$ is defined as $E[u] = \int\limits_X u dP$. How does this equal $\int\limits_{\mathbb{R}} x f(x) dx$ where $f(x) = \frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}$?

I guess that the induced measure has to be brought into play along with some Radon-Nikodym theorem to show this but I am not sure how to do so.

Any help is greatly appreciated.

Thanks, Phanindra

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up vote 8 down vote accepted

The definition of induced measure can be expressed as equality of integrals over the two probability spaces $$\int_\Omega 1_B(u(\omega))\, P(d\omega) = \int_\mathbb{R} 1_B(x)\, Q(dx).$$

This equation is true for every Borel set $B$, and can be extended to all bounded Borel measurable functions $g$; that is, $$\int_\Omega g(u(\omega))\, P(d\omega) = \int_\mathbb{R} g(x)\, Q(dx).$$

Provided the mean of $u$ exists, you can use truncation and dominated convergence to extend this to the unbounded function $g(x)=x$ and conclude that $$\int_\Omega u(\omega)\, P(d\omega) = \int_\mathbb{R} x\, Q(dx).$$

If the random variable $u$ has a density function $f$, then integration with respect to $Q(dx)$ can be replaced by integration with respect to $f(x)\,dx$ so that the mean is $E(u)=\int_\mathbb{R} x f(x)\,dx$.

This fact is sometimes expressed with the funny terminology: the law of the unconscious statistician.

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Byron: Thanks for the answer. It has given me a direction to think in. I believe I need to know more to understand your answer. I am a bit pleasantly surprised to have gotten a reply from a Prof at my alma mater. –  jpv Oct 6 '11 at 18:24
    
@jpv Great! Glad to be of help. –  Byron Schmuland Oct 6 '11 at 18:28
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