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I've been trying to understand this question I have below and I've been using this as a reference.

The question:

I'm given a random variable

$X(\omega)=\begin{cases} \omega, \quad 0\leq\omega\leq \frac{1}{3}\\ \frac{1}{3}, \quad \frac{1}{3}<\omega\leq 1 \end{cases}$

I'm looking for its CDF and the probability measure is defined as $P([a,b])=b-a$.

The above example (linked post) is very helpful as it illustrates it using a uniform distribution, but I suppose I'm a bit confused on the part of $X(\omega)=\frac{1}{3}$ when $\frac{1}{3}<\omega \leq 1$.

So far my CDF looks like: $P(X\leq x)=x\mathbb{1}_{[0,\frac{1}{3}]}$ but I'm not sure how to the CDF works from $(\frac{1}{3},1]$.

Idrees
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    Please add `self-study` as a tag to avoid others mistakenly providing the complete resolution rather than help into solving the exercise. – Xi'an May 25 '20 at 05:42
  • You're right about $F_X(x) = x,$ for $0 1.$ Next, $P(X=1/3) = 2/3,$ so the CDF jumps from $1/3$ to $1$ at $x = 1/3.$ // Remember that a CDF is nondecreasing. What is its value at $x = .5?$// From 0 to 1/3, X acts like a continuous RV, then it acts like a discrete RV with probability 2/3 at 1/3. The CDF of a strictly continuous RV has no jumps. // The CDF of the discrete random variable $Y \sim \mathsf{Binom}(n = 2, p = 1/2)$ has jumps at 0, 1, and 2 of sizes 1/4, 1/2, 1/4, respectively; a 'pure jump' function. – BruceET May 25 '20 at 07:14
  • The method of *drawing a picture* (of the graph of $X$) illustrated at https://stats.stackexchange.com/questions/138763 makes short work of this problem. – whuber May 26 '20 at 13:19

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