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Hey I tried searching online but I can't find this proof based on the CDF.

$$P(a < X \le b) = F(b) - F(a)$$

Thanks for any help.

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I think it should be clear if you write out what $F(b)$ and $F(a)$ are. –  eeeeeeeeee Apr 28 '13 at 4:48
I mean, there isn't really much to prove. –  eeeeeeeeee Apr 28 '13 at 4:48
@wzsun Which parts of the accepted answer were you unaware of when you asked the question? –  Did Apr 28 '13 at 6:47
I must admit, I have always found probability to feel a bit unnatural.... it requires a lot of different thinking. And for me, I was confused at the start of reading Rosenthal because PDF is just called distribution function. –  Squirtle Apr 29 '13 at 5:13

1 Answer 1

up vote 1 down vote accepted

By definition the CDF (cumulative density/distribution function) is computed by taking a "cumulative" integral up to the given point of the PDF $(f_X(x)):$ $$F(a):=P(X \le a):=\int_{-\infty}^a f_X(x)dx=1-P(a<X)=1-\int_{a}^\infty f_X(x) dx$$

I included the part towards the right because I thought it might help you 'see' what's going on.

Likewise $$P(X\le b):=\int_{-\infty}^b f_X(x)dx=F(b)$$

So $$P(\{a\le X \} \cap \{X\le b\})=P(a\le X \le b) = \int_{-\infty}^b f_X(x)dx - \int_{-\infty}^af_X(x)dx=F(b)-F(a)$$

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