I am having trouble understanding how to get a complete mixed distribution function from a mixed CDF. The question is from my actuarial exam p manual. Namely, the discrete portion is confusing me.
The question is:
The claim amount random variable B has the following distribution function
F(x)
={ $0 , x<0 \ \ ;\frac{x}{2000},0\leq x<1000 \ ; \ \ .75, x=1000 \ ; \ \ \frac{x+11,000}{16,000}, 1000<x<5000 \ ; \ \ 1, x\geq5000 \} $
I can see that f(x)= $\{0,x<0 \ \ ; \ \ \frac{1}{2000}, 0 \leq x <1000 \ ; \ \ c,x=1000 \ ; \ \ \frac{1}{16000}, 1000<x<5000 \ ; \ \ 0, x \geq 5000 \}$
and then clearly you must sum up all these components to equal 1. ie: $1=999(\frac{1}{2000})+c+3998(\frac{1}{16000}) \ \ \ \Longrightarrow c=0.749375$.
My question is, what would we do if there were two discrete oarts. How would we find the whole mixed probability distribution function.
By the definition of distribution function, $F(x)=P(X\le x)=P(X<x)+P(X=x)$. For a continuous distribution, $P(X=x)=0$ so that $F(x)=F(x^-)$. In your case, however, $$F(1000)=0.75\ne F(1000^-)=\lim_{\varepsilon\to0}\frac{1000-\varepsilon}{2000}=0.50$$This indicates that $P(x=1000)\ne0$ and the distribution is a mixture of both continuous and discrete types. To find the value of $P(x=1000)$, simply note$$F(x)-P(X<x)=F(x)-F(x^-)=P(X=x)$$giving $P(x=1000)=0.75-0.50=0.25$. Thus, a discontinuity in the distribution function indicates discrete points of non-zero probability.