If we know that $P_X=\int f_X \ d\mu$, where $\mu$ is another measure(could be lebesgue) and I know what I get from $P_X(B), \ \forall B\in \mathcal{B}(\mathbb{R})$, how can I deduce the function $f_X$? Any help would be appreciated.

Edit: Some comments have suggested $\frac{d P_X(]-\infty,x])}{dx}=f(x)$, using the fundamental theorem of calculus. However, the only version I know is the following:

If $f:[a,b]\rightarrow \mathbb{R}$ is continuous, then $f$ is integrable, and the function $F$ given by $F(x)=\int^x_a f \ dm$ is differentiable for $x \in (a,b)$ with derivative $F'=f$. The measure $m$ is Lebesgue.

We usually define $F(x)=P_X(]-\infty,x])$. So, what changes do we need to do to the text of the above theorem to be able to apply it to the case $F(x)=P_X(]-\infty,x])$?

  • $\begingroup$ By $P_X(B)$, do you mean $\int_Bf_Xd\mu$? $\endgroup$ – Michael M Jan 5 '16 at 16:15
  • $\begingroup$ @MichaelM. Yes, I do. $\endgroup$ – An old man in the sea. Jan 5 '16 at 16:18
  • 2
    $\begingroup$ $$f_X(x)=\frac{d}{dx}P_X((-\infty,x])$$ $\endgroup$ – Did Jan 5 '16 at 16:21
  • $\begingroup$ @Did, I thought that theorem was only valid when $P_X((-\infty,x])=P_X([a,x])$. The way it's written in the notes/textbook I'm using is that $F(x)=\int_a^xf_X \ d \mu$ Thanks for the helpful comment $\endgroup$ – An old man in the sea. Jan 5 '16 at 16:27
  • 1
    $\begingroup$ If the $-\infty$ bound annoys you, fix some $a$ and note that, for every $x$, $$F(x)=c+\int_a^xf(t)dt,\qquad\text{with}\ c=\int_{-\infty}^af(t)dt.$$ $\endgroup$ – Did Jan 7 '16 at 17:36

Formally you can use

$$ f(x) = \mathsf{E}[ \delta(X-x)] $$

where $X$ is the random variable, $x$ the values it can attain, and $\mathsf{E}$ the expectation value. Alternatively you can compute the cdf:

$$ cdf(x) = \mathrm{Prob}(X < x) = \mathsf{E}[ \theta(x-X)] $$

where $\theta$ is the Heaviside function. You can then obtain $f(x)$ by differentiating the $cdf$.

  • 1
    $\begingroup$ Why point to difficult-to-master physicists "definitions" when it is so simple to write down the true definitions? $\endgroup$ – Did Jan 5 '16 at 16:34
  • $\begingroup$ This could open up a long discussion. Dirac deltas have been used long before "their true definition" (whatever that means) was given. I believe that the difficult-to-master physicists "definitions" is actually very illuminating (and this was OP's question). Moreover you know very well that what I wrote can be perfectly made sense of. $\endgroup$ – lcv Jan 5 '16 at 16:44
  • $\begingroup$ @Did In any case I started my answer with the disclaimer "formally". The second part of my answer is the same as yours. $\endgroup$ – lcv Jan 5 '16 at 16:50
  • $\begingroup$ "True definition" means "mathematically rigorous definition", there is no mystery involved. Of course "what (you) wrote can be perfectly made sense of", unfortunately this fact is perfectly offtopic here. $\endgroup$ – Did Jan 5 '16 at 16:51
  • 1
    $\begingroup$ @Did Lastly: even "your" answer via "differentiation" is only correct in those circumstances where standard differentiation makes sense, otherwise a more general definition of differentiation is needed. You are using a notation too. $\endgroup$ – lcv Jan 5 '16 at 17:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.