# From measure $P_X$ and knowing that $P_X=\int f_X d\mu$, how to find $f_X$?

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 $F(x)=P_X(]-\infty,x])$?

• By $P_X(B)$, do you mean $\int_Bf_Xd\mu$? – Michael M Jan 5 '16 at 16:15
• @MichaelM. Yes, I do. – An old man in the sea. Jan 5 '16 at 16:18
• $$f_X(x)=\frac{d}{dx}P_X((-\infty,x])$$ – Did Jan 5 '16 at 16:21
• @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 – An old man in the sea. Jan 5 '16 at 16:27
• If the $-\infty$ bound annoys you, fix some $a$ and note that, for every $x$, $$F(x)=c+\int_a^xf(t)dt,\qquad\text{with}\ c=\int_{-\infty}^af(t)dt.$$ – 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 $cdf$.

• Why point to difficult-to-master physicists "definitions" when it is so simple to write down the true definitions? – Did Jan 5 '16 at 16:34
• 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. – lcv Jan 5 '16 at 16:44
• @Did In any case I started my answer with the disclaimer "formally". The second part of my answer is the same as yours. – lcv Jan 5 '16 at 16:50
• "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. – Did Jan 5 '16 at 16:51
• @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. – lcv Jan 5 '16 at 17:13