What would the intuitive explanation be?
Would it be because there are infinately many values and so the probability of any specific value is infinitely small hence we say 'close enough' to 0?
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3If it were nonzero, then the distribution would have a point mass there and so not be a continuous distribution. You can think of it as $P_X(x_0)=\lim\limits_{\epsilon\to 0}\int\limits_{x_0-\epsilon}^{x_0+\epsilon}f_X(x);dx = 0$ – MPW May 22 '19 at 20:36
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A continuous random variable has the following property
$$ P( a \leq x \leq b) = \int_{a}^{b} f(x) \textrm{d}x $$
where the pdf of the RV is given by $f(x)$.
In calculus we know that if the upper and lower limits of the integral are the same then it is $0$.
$$ P( c \leq x \leq c) = \int_{c}^{c} f(x) \textrm{d}x = 0 $$
You could probably justify this a few ways, but from the fundamental theorem of calculus we have that
$$ F(b) - F(a) = \int_{a}^{b} f(x) \textrm{d}x $$
so then the integral at a single point is
$$ F(c) - F(c) = 0 $$
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I +1d but, not all continuous random variables have PDFs. Of course, practical ones do, and so it is good intuition. – Michael May 22 '19 at 23:19
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In continuous case probabilities are defined as integrals. The probability of a single point is zero because you have an integral whose upper and lower limits are the same.
The intuitive explanation is that you have uncountable infinitely many options with and you randomly pick one of them. What is the probability of choosing a specific one?
The randomness among uncountable infinite number of choices makes it very unlikely to come up with the one that you like to choose.
If you randomly pick a person what is the probability that he or she has exactly the same height as you have?
Mohammad Riazi-Kermani
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I +1d but, we can have an infinite number of choices $i \in {1, 2, 3, ...}$, each choice with positive probability $P[{i}] = 1/2^i$ for $i \in {1, 2, 3, ...}$. – Michael May 22 '19 at 22:29
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Nevertheless your argument holds if you modify to: "you have infinitely many options [each with equal probability] and you randomly pick one of them. What is the probability of choosing a specific one?" – Michael May 22 '19 at 23:03
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Thanks for the update! I have deleted my comment on the first paragraph that is no longer relevant. – Michael May 22 '19 at 23:47
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Intuition
(Building on the MPW comment): If $P[X=x^*]=p>0$ at a certain point $x^*$ then \begin{align} &\{X\leq x^*\} = \{X=x^*\} \cup \{X<x^*\}\\ &\\ \implies \quad &P[X\leq x^*] = p+P[X<x^*] \quad \quad (Eq. 1) \end{align} You can see there is a positive gap of size $p$ between $P[X\leq x^*]$ and $P[X<x^*]$, which intuitively shows why the CDF function $P[X\leq x]$ jumps discontinuously at $x^*$.
Thus, if the CDF has no discontinuous jumps then there can be no such point masses and so $P[X=x]=0$ for all $x \in \mathbb{R}$.
A full proof
Assume $P[X=x^*]=p>0$. Fix $x<x^*$ and notice that $$ \{X\leq x\} \subseteq \{X < x^*\} \implies P[X\leq x] \leq P[X< x^*]$$ Substituting this inequality into (Eq. 1) gives $$ P[X\leq x^*] \geq p+ P[X\leq x] \quad \forall x < x^*$$ This holds for all $x<x^*$ and so it holds even when taking a limit as $x$ approaches $x^*$ from the left: $$ P[X\leq x^*] \geq p+\lim_{x\nearrow x^*}P[X\leq x] $$ So there is a positive gap of size (at least) $p$ between the value $P[X\leq x^*]$ and the limit of the CDF function $P[X\leq x]$ as $x$ approaches $x^*$ from the left.
*In fact it can be shown that $\lim_{x\nearrow x^*}P[X\leq x]=P[X<x^*]$ and so from (Eq. 1) we see the gap is exactly $p$. The size of the discontinuous jump in the CDF function at $x^*$ is exactly the probability that $X=x^*$.
Michael
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The above proof that the gap is at least $p$ holds even for finitely additive spaces. The footnote fact that the gap is exactly $p$ is proven via the countable additivity axiom. I believe the gap can be larger than $p$ in spaces that are finitely additive but not countably additive (which are not probability spaces). Somehow I am having trouble coming up with an example of such, though. – Michael May 22 '19 at 23:11
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Okay, to organize my own thoughts on my above comment: define a finitely additive (but not countably additive) probability measure on $\mathbb{N}={1, 2, 3, …}$ via $P:2^{\mathbb{N}}\rightarrow\mathbb{R}$ where $A \subseteq \mathbb{N}$ defines $P[A]$ as a Banach limit of ${\frac{|A \cap {1, 2, …, k}|}{k}}_{k=1}^{\infty}$ as in my answer here, which has $P[A]=0$ for every finite set $A$: https://math.stackexchange.com/questions/204842/example-for-fintely-additive-but-not-countably-additive-probability-measure – Michael May 23 '19 at 00:37
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Now define $X:\mathbb{N}\rightarrow\mathbb{R}$ by $$ X(n) = \left{ \begin{array}{ll} 1-1/n &\mbox{ if $n$ is odd} \ 1 & \mbox{ if $n$ is even} \end{array} \right.$$ Then for all $\delta>0$ we have \begin{align}P[X\leq 1]&=P[\mathbb{N}]=1\P[X=1]&=P[{\mbox{even natural numbers}}]=1/2\ P[X\leq 1-\delta]&=P[{\mbox{odd natural numbers less than or equal to $1/\delta$}}]=0\end{align} So the CDF function $P[X\leq x]$ has a discontinuous jump of size $1$ at $x=1$, but $P[X=1]=1/2$ so the point mass is smaller than the jump! – Michael May 23 '19 at 00:44