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I'm trying to better understand the measure theoretic definition of a probability density function, but I believe I'm making a mistake somewhere if someone could clarify. From what I currently gather, let $(\Omega,F,\mathbb{P})$ be a probability space and let $X:\Omega\rightarrow \mathbb{R}$ be a real valued random variable. Then, the distribution of $X$ is the push-forward measure $$ X_*\mathbb{P}(E)=\mathbb{P}(X^{-1}(E))=\mathbb{P}(X\in E) $$ where $E\subseteq \mathbb{R}$ is some measurable set in $\mathbb{R}$. The cumulative distribution function of $X$ is then $$ F(a) = \mathbb{P}(X\in (-\infty,a]) = \mathbb{P}(X^{-1}(-\infty,a]) $$ To obtain the probability density function, we assume that $\mathbb{P}<<\lambda$ where $\lambda$ is the Lebesgue measure and then note $$ F(a)=\mathbb{P}(X^{-1}(-\infty,a]) = \int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} = \int_{X^{-1}(-\infty,a]} \frac{d\mathbb{P}}{d \lambda} d\lambda $$ where $\frac{d\mathbb{P}}{d \lambda}$ is the Radon-Nikodym derivative of $\mathbb{P}$ with respect to $\lambda$. We then set $f=\frac{d\mathbb{P}}{d \lambda}$ and call $f$ the probability density function.

Alright, so I know there's an error here since I really want $$ F(a) = \int_{-\infty}^a f d\lambda $$ but instead I have $$ \int_{X^{-1}(-\infty,a]} f d\lambda $$ Basically, there's $X^{-1}$ here and I don't think there should be, so what's the error in what I'm doing with Radon-Nikodym? Second, I'm still not entirely sure how to define $\mathbb{P}$ concretely. Say we're working with a normal distribution, I know that $$ f(x) = \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x-\mu)^2}{2\sigma^2}} $$ and $$ F(x) = \frac{1}{2}\left[1+\textrm{erf}\left(\frac{x-\mu}{\sigma\sqrt{2}}\right)\right] $$ In this case, what would $\mathbb{P}$ be?

Thanks for the help in advance.

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  • $\begingroup$ This might be helpful: en.wikipedia.org/wiki/Probability_density_function $\endgroup$ – i707107 Oct 8 '16 at 5:17
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    $\begingroup$ Try to spot which measure is defined on which set: here, $X:\Omega\to\mathbb R$, $P$ is a measure on $\Omega$ and $\lambda$ is a measure on $\mathbb R$, hence $P\ll\lambda$ is absurd and should read $X_*P\ll\lambda$. The rest follows, for example, the PDF of $X$, when it exists, is $$f=\frac{d(X_*P)}{d\lambda}$$ which is equivalent to the fact that for every Borel subset of $\mathbb R$, $$P(X\in B)=\int_Bf\,d\lambda$$ $\endgroup$ – Did Oct 8 '16 at 8:17
  • $\begingroup$ @Did Makes sense and that helps a lot. Could you clarify the second part? Basically, for a normal distribution, how do we define the measure $P$ on $\Omega$? $\endgroup$ – wyer33 Oct 9 '16 at 4:41
  • $\begingroup$ Again: the point is that one does not need to "define P concretely" since P lives on the source set Omega while one works with the PDF and the CDF, that are defined on the image set. $\endgroup$ – Did Oct 9 '16 at 6:07
  • $\begingroup$ @Did Certainly, but if we wanted to define it, could we? Alternatively, am I going about this backwards? Meaning, start with $$ F(B)=\int_B f d\lambda $$ then, push $f$ into the measure $$ F(B)=\int_B 1 d\mu $$ using Radon-Nikodym where $(\mathbb{R},B,\mu)$. Then, for some random variable $X:\Omega\rightarrow\mathbb{R}$, there exists a probability measure $P$, $(\Omega,F,P)$, such that $\mu=X_*P$. We may not know what $P$ is and we may not need it, but one exists. $\endgroup$ – wyer33 Oct 9 '16 at 20:21
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This amounts to proving that

$P[X\in B]=\int_BdF,$

where $B$ is a Borel set in $(-\infty, \infty)$ and $X$ is a random variable with distribution funtion $F$.

Let $\mu_F$ denote the Lebesgue-Stieltjes measure over $(-\infty,\infty)$ determined by $F$, and let's take $B$ of the form $B=\cup_{i=1}^n(a_i,b_i]$. Then we have

$\mu_F(B)=\sum_{i=1}^n(F(b_i)-F(a_i))=\sum_{i=1}^nP(a_i< X \leq b_i)=P(\cup_{i=1}^n\{a_i < X \leq b_i\})=P(X^{-1}B)$

$PX^{-1}$ is a measure since $\mu_F$ is, hence we can extend uniquely both measures to the sigma field of borel sets in $\mathbb{R}$.

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You claim the following change of variables: $$\int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} \overset{?}{~=~} \int_{\color{crimson}{X^{-1}(-\infty,a]}} \frac{d\mathbb{P}}{d \lambda} d\lambda$$

However, you should change the integration interval when you changed the variable.

$$\int_{X^{-1}(-\infty,a]} 1 d\mathbb{P} ~=~ \int_{\color{navy}{(-\infty,a]}} \frac{d\mathbb{P}}{d \lambda} d\lambda$$

That is all.

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