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Let be $X$ an absolutely continuous random variable. Then, we know that there exists a probability density function (pdf) $f_X$ such that its cumulative distribution function (cdf) $F_X(x)$ can be represented by $$ F_X(x)=\int\limits_{-\infty}^xf_X(t)dt. $$

In this context our professor made the statement, that

$F_X(x)$ is continuously differentiable $\iff$ $f_X$ is continuous.

I tried to verify both directions:

"$\impliedby$":

If $f_X$ is continuous, then the statement follows from the Fundamental Theorem of Calculus (applied to improper integrals).

"$\implies$":

Maybe be we can look at $\int\limits_{-\infty}^xF'_X(t)dt$, which also is an antiderivative of $F'_X(x)$ and compare it to $F_X(x)=\int\limits_{-\infty}^x f_X(t)dt$. My guess was that $F'_X(x)$ could be utilized as pdf which would lead to something like $F_X(x)=\int\limits_{-\infty}^x f_X(t)dt=\int\limits_{-\infty}^x F'_X(t)dt$. But I am not sure if this is the way to go...

Any help is welcome!

PS: I know that there already exist similar questions but mostly they are answered wrong, rely on additional assumptions or leave out important details, see for example https://math.stackexchange.com/a/248272/579544

Philipp
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1 Answers1

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Continuously differentiable functions satisfy FTC so $F_X(y)-F_X(x)=\int_x^{y}F_X'(t)dt$ for $x <y$. $F_X'(t)$ is non-negtaive because $F_X$ is increasing. Letting $x \to -\infty$ and $y \to +\infty$ in $F_X(y)-F_X(x)=\int_x^{y}F_X'(t)dt$ we see that $F_x'$ integrates to $1$. Hence, $F_x'$ is a density function. Since $P(a<X\leq b)=\int_a^{b} F_X'(t)dt$ it follows that $P(X\in A)=\int_A F_X'(t)dt$ for every Borel set $A$ and $F_X'(t)dt$ is indeed the density function of $X$. Thus, $f_X=F_X'$ which is continuous.

geetha290krm
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  • You have only shown that $F'_X$ is another pdf of $X$. I don't see how you can conclude $F'_X=f_X$. – Philipp Mar 23 '23 at 15:01
  • A r.v. can have only one pdf (up to) a set of measure $0$. @Philipp – geetha290krm Mar 23 '23 at 15:06
  • Not sure what arv means, but I assume in order to understand and apply your reasoning one needs more knowledge about measure theory. Our professor made the original statement during a introductory class of stochastics where measure theory is not a prerequiste. So maybe there is a chance of proving the statement in a more elementary way? – Philipp Mar 23 '23 at 15:15
  • You have given a definition of $f_X$. But do you know that $f_X$ in the definition is unique? Otherwise, the definition does not even make sense. I am using th fact that there is only one $f_X$ such that $F_X(x)=\int_{-\infty}^{x} f_X(t)dt$ for all $x$. [BTW, A r.v. means A random variable]. @Philipp – geetha290krm Mar 23 '23 at 23:12
  • It's just the textbook definition of an absoutely continuous random variable which of course doesn't say that $f_X$ is unique. Not sure what you are referring to by "the definition does not even make sense"? I guess the statement from our professor is a bit misleading if not wrong, it should be instead: "$F_X(x)$ is continuously differentiable" $\iff$ "there exists some continuous probability density function". – Philipp Mar 24 '23 at 11:27
  • If uniqueness is not within the results covered so far, then yes, you can only say there exists some continuous probability density function. @Philipp – geetha290krm Mar 24 '23 at 11:31