For random variable $X$ that follows some distribution, $f(x)$ is the probability density function of that distribution if and only if $$\mathbb{E}[\phi(X)] = \int_{-\infty}^\infty \phi(x) f(x)dx$$ for all functions $\phi$.

Context: My professor used this in lecture to demonstrate a way to find the distribution of $cX$ given random variable $X$ that follows a specific distribution.

  1. What is the name of this theorem?
  2. Have I missed any qualifications/caveats? I am in particular curious about whether "for all functions" is correct. My professor mentioned "positive, bounded, deterministic functions," but I am not sure what he meant by that.
  3. Where can I find a proof of this theorem?
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    $\begingroup$ I don't think this is a theorem; I think it's a definition of "probability density function." $\endgroup$ – Qiaochu Yuan Nov 24 '12 at 4:01
  • $\begingroup$ Ah I see... just to clarify, is the following statement correct? If I show that the equation is true for one function phi, it is not enough to say that f(x) is the pdf of the distribution...? $\endgroup$ – angryavian Nov 24 '12 at 4:23
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    $\begingroup$ My lecturer used to call this the 'Law of the unconscious statsitician'. Not sure how standard that is though. $\endgroup$ – John Gowers Feb 28 '15 at 23:46
  • $\begingroup$ @Donkey_2009 Yes, this is how Ross cites it, if I recall. $\endgroup$ – Clarinetist May 17 '15 at 18:46

A probability density function is usually defined in the following way:

Let $X$ be a random variable. Then $f$ is the probability density function of the distribution given by $X$ if and only if $f(x)\geq 0$ for all $x\in\mathbb{R}$ and $$ P(X\in A)=\int_A f(x)\,\lambda(\mathrm d x), \quad A\in\mathcal{B}(\mathbb{R}). \tag{1} $$

However, $(1)$ is equivalent to both of the following

$$ P(a\leq X\leq b)=\int_a^bf(x)\,\lambda(\mathrm dx),\quad -\infty<a<b<\infty.\tag{2} $$


$$ \mathrm{E}[\varphi(X)]=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx),\tag{3} $$ for every measurable and integrable (with respect to the measure $f\lambda$) function $\varphi$.

It is clear that $(3) \Rightarrow (2)$ (take $\varphi=1_A$) and $(1)\Rightarrow (2)$ (take $A=[a,b]$). But on the other hand $(2)\Rightarrow (1)$ by the use of Dynkin's lemma and also $(1)\Rightarrow (3)$ by a standard argument that is often used in probability theory. The argument goes as follows:

a) The property $(3)$ holds for all indicator functions according to $(1)$.

b) If $\varphi$ and $\psi$ are two functions satisfying $(3)$, then $$ \begin{align*} \mathrm{E}[(\varphi+\psi)(X)]=\mathrm{E}[\varphi(X)]+\mathrm{E}[\psi(X)]&=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx)+\int_\mathbb{R}\psi(x)f(x)\,\lambda(\mathrm dx)\\ &=\int_\mathbb{R}(\varphi+\psi)(x)f(x)\,\lambda(\mathrm dx), \end{align*} $$ and so $\varphi+\psi$ satisfies $(3)$. In a similar fashion one can show that $\mathrm{E}[\alpha \varphi(X)]=\alpha\mathrm{E}[\varphi(X)]$ for $\alpha\in\mathbb{R}$ and hence the set of functions satisfying $(3)$ form a vector space.

c) Suppose that $(\varphi_n)_{n\geq 1}$ is a sequence of non-negative, increasing functions satisfying $(3)$ such that $\varphi=\lim_{n\to\infty}\varphi_n$ exists pointwise. Then by applying the monotone convergence theorem (twice) we have $$ \mathrm{E}[\varphi(X)]=\lim_{n\to\infty}\mathrm{E}[\varphi_n(X)]=\lim_{n\to\infty}\int_\mathbb{R}\varphi_n(x)f(x)\,\lambda(\mathrm dx)=\int_\mathbb{R}\varphi(x)f(x)\,\lambda(\mathrm dx). $$

Now the standard argument yields that $(3)$ holds for every measurable and integrable (with respect to the measure $f\lambda$) $\varphi$.

This also shows that $\varphi$ in $(3)$ could just as well have been chosen to be "positive and bounded".

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    $\begingroup$ +1. But replace measurable and integrable in $(\ast\ast\ast)$ by measurable and integrable with respect to the measure $f\lambda$ or an equivalent formulation. $\endgroup$ – Did Nov 24 '12 at 10:54
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    $\begingroup$ @did: Thanks. Edited accordingly. $\endgroup$ – Stefan Hansen Nov 24 '12 at 11:00
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    $\begingroup$ +1. Just curious: don't you need $\mathbb{E}c\phi\left(X\right)=c\mathbb{E}\phi\left(X\right)$ also? $\endgroup$ – drhab May 17 '15 at 18:17
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    $\begingroup$ Yeah, you're right @drhab. Thanks for pointing that out. $\endgroup$ – Stefan Hansen May 17 '15 at 18:41

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