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I am trying to understand how to do these simple questions where one has to find probability, given pdf or cdf.

Here is the question:

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This one is a cdf and from my understanding of a cdf, $P(2\le X<4)$ should be equal to $F(4)-F(2)$ which comes out to $0.25$ but answer given is $0.4$.

What am I missing?

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    $P(2≤X<4) =1 -P(X<2)$. But $P(<2)=.6$ so this is $.4$. – 2017-02-04
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    $F(2)=1/2+2/8=0.75$???? – 2017-02-04
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    Yes, I edited. I assumed (without looking) that the distribution was continuous. It isn't. You don't want $F(2)$ you want the probability that $X<2$. For continuous distributions this is the same, but for discrete distributions it needn't be. – 2017-02-04
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    why is it discrete? – 2017-02-04
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    Because the cdf is discontinuous. $\lim_{x\to 2^-}F(x)=.6\neq F(2)$. Thus there is a non zero probability that $X=2$. Specifically, $P(X=2)=.75-.6=.15$. – 2017-02-04
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    @lulu ok so i get it now i guess. If cdf is discontinuos at any break points, which this one is at 1 and 2, $F(x_0)\neq P(X\le x_0)$. But in this non continuos cases how to find probablities, can you explain plz – 2017-02-04
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    Note: the distribution isn't purely discrete.it is continuous between $0$ and $!$ and again from $2$ up. But there is a non-zero probability that $X=1$ and a non-zero probability that $X=2$. – 2017-02-04
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    Here it's easy because you can write out the probabilities exactly. In general, to get $P(X<2)$ you'd need to compute $\lim_{x\to 2^-}F(x)$ which might be hard if $F(x)$ is complex. But here we can see that there is no chance that $1 – 2017-02-04
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    To use different words: Since we can't have $1 – 2017-02-04

1 Answers 1

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We can drop "$X < 4$" in $P(2 \le X < 4)$ since $F(x) = P(X \le x) = 1 \forall x \ge 3$. (In particular, taking $x = \pi$ gives $F(\pi) = P(X \le \pi) = 1$. Translating this into English gives "$X$ is less than or equal to $\pi$ (i.e. 4) almost surely".) Writing the probability in the set theoretic definition, we have \begin{align} & P(X < 2) \\ =& P(\{\omega\in\Omega\mid X(\omega)<2\}) \\ =& \sup_{\epsilon>0} P(\{\omega\in\Omega\mid X(\omega)\le2-\epsilon\}) \\ =& \sup_{0<\epsilon<1} P(\{\omega\in\Omega\mid X(\omega)\le2-\epsilon\}) \\ =& \sup_{0<\epsilon<1} F(2-\epsilon) \\ =& \sup_{1

So $P(2\le X<4) = P(X \ge 2) = 1 - P(X>2) = 1 - \frac35 = \frac25$, which is OP's given answer.