Evaluating mass of probability for a single point when the distribution is truncated

Question

I have a problem with evaluating the probability mass at $K$ for truncated exponential disribution:

$$ F(t)= \begin{cases} 0,& t<0\\ 1-e^{-\lambda t}, & 0\leq t<K\\ 1, & t\geq K \end{cases} $$

How to find the probability mass at a single point in $K$? In the lecture it is written that it should be $e^{-\lambda K}$. But why? Thanks in advance.

Well you have the mixed density $f(t) = \lambda e^{-\lambda x}\cdot\mathsf 1_{(0,K)}+ (1-e^{-\lambda K})\cdot\delta_K$. But the probability $\mathbb P(X=x)$ is zero when $x\ne K$. — Math1000, Oct 05 '19 at 15:00
@callculus When I'm differentiating $1$ (because it's distribution for $K$), it gives me $0$ what doesn't make sense when I'm looking at answer. — Novice, Oct 05 '19 at 15:05
@Math1000 my question focuses on situation when $t$ is exactly $K$. — Novice, Oct 05 '19 at 15:11
There's a point mass at $x=K$. $\Pr(X\leq K)=1$ but $\lim_{x\to K^-}\Pr(X\leq x)=1-e^{-\lambda K}$ So, $\Pr(X=K)=1-(1-e^{-\lambda K})=e^{-\lambda K}$ — saulspatz, Oct 05 '19 at 15:17

score 2 · Accepted Answer · answered Oct 05 '19 at 15:16

2

$$P(X=K)=P(X\leq K)-P(X<K)$$

This with $P(X\leq K)=F(K)=1$ and: $$P(X<K)=\lim_{t\uparrow K}P(X\leq t)=\lim_{t\uparrow K}F(t)=1-e^{-\lambda K}$$

answered Oct 05 '19 at 15:16

drhab

151,093

Evaluating mass of probability for a single point when the distribution is truncated

1 Answers1