Contradiction when derive pdf from cdf and integrate again

Question

I encountered this example on statlect.com and the screenshot of it is as follows

It took me a while to figure out why $P(X=1)=1/2$. I don't have much experience dealing with distribution functions with both discrete and continuous parts. So I am not sure if it is legal to conclude contradiction does arise when doing the following operations:

PDF of $X$ is $f(x) = 1/2$ when $x \in (1,2)$. In addition $f(1)=1/2$. Therefore $f(x)$ is continuous on $[1,2)$. That makes $\int_{1}^{2}f(x)=\frac{1}{2}$. However, that would make $f(x)$ not a legitimate PDF and it would only make sense to do $\int_1^{2}f(x)+f(1) = 1$. But does it mean we have different results when we integrate $f(x)$ on $(1,2]$ and $[1,2]$? How can this happen?

Some digression: I am actually not sure how $f(x)$ should behave when $x=2$ since $F(x)$ can not be differentiated at that point.

Answer 1

Note that $$f(x) = \frac{d}{dx}F(x).$$ Consequently, we have $$f(1) = \frac{1}{2}\delta(x),$$ where $\delta(x)$ is the Dirac delta function. Also, $$\int_{1^-}^{2} f(x) dx = \int_{1^-}^{1^+} \frac{\delta(x)}{2}dx + \int_{1^{+}}^{2} f(x) dx = \frac{1}{2} + \frac{1}{2}=1.$$

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For more information, check out this website about mixed random variables.

Answer 2

Note that in general $$ P(X=a)=P(X\le a)-P(X<a)=F(a)-\lim_{x\uparrow a}F(x). $$ In your example $$ P(X=1)=F(1)-\lim_{x\uparrow 1}F(x)=\frac{1}{2}-0=\frac{1}{2}. $$ To see why $\lim_{x\uparrow a} F(x)=P(X<a)$, observe that $$ (X\leq a_{n})\uparrow (X< a)\implies \lim_{x\uparrow a}F(x)=\lim_{n\to \infty} P(X\leq a_n)=P(X<a) $$ by measure continuity from below where $(a_n)$ is strictly increasing with limit $a$.

Also $X$ cannot have a density function with respect to Lebesgue measure since $P(X=1)>0$. Integrating a density over a single point must yield zero.