In uniform distribution, a continuous distribution, for example where $A = -1$ and $B = 1$, $P(X < 0)$ is said to be the same as $P(X \le 0)$. Why is this?
2026-04-03 23:00:20.1775257220
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Why is $P(X<0)$ the same as $P(X\le 0)$ for continuous distributions?
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I don't know what you mean with $A,B$ but I suppose that with a continuous distribution you mean a distribution on the real line which has a density, i.e. $$ P(X\in [a,b]) = \int\limits_a^bf(x)\,dx. $$ Note that $$ P(X\leq 0) = P(X<0)+P(X=0) = P(X<0)+\int\limits_{0}^0f(x)dx $$ and the latter integral is zero as Jonas has written.
Generally, continuous distribution means $P(X=c)=0$ for every $c$. Thus in your case $P(X=0)=0$, and this is the difference between the two probabilities you mention.
Note, though, that continuous does not imply absolutely continuous, so the distribution may not be of the form $P(X \in A) = \int_A f(x) dx$ for any function $f$. The word continuous comes from writing the distribution as $P(X \in A) = \int_A \,dF(x)$ in Stieltjes form, and requiring $F$ to be continuous.