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Exercise 11.2 · Q5

Q.The cumulative distribution function of a discrete random variable is given by
[!FORMULA] F(x)={0−∞<x<−10.15−1≤x<00.350≤x<10.601≤x<20.852≤x<313≤x<∞F(x)=\begin{cases}0 & -\infty<x<-1\\ 0.15 & -1\le x<0\\ 0.35 & 0\le x<1\\ 0.60 & 1\le x<2\\ 0.85 & 2\le x<3\\ 1 & 3\le x<\infty\end{cases}
Find

(i) the probability mass function
(ii) P(X<1)P(X<1) and
(iii) P(X≥2)P(X\ge2).
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XX's support is exactly the jump points of FF: −1,0,1,2,3-1,0,1,2,3; each jump size F(xi)−F(xi−1)F(x_i)-F(x_{i-1}) is the pmf value there, and the two probabilities follow directly from the given cdf table.

Step 1. (i) Read off the pmf as jump sizes.

f(−1)=F(−1)−F(−1−)=0.15−0=0.15f(-1)=F(-1)-F(-1^-)=0.15-0=0.15.

f(0)=F(0)−F(−1)=0.35−0.15=0.20f(0)=F(0)-F(-1)=0.35-0.15=0.20.

f(1)=F(1)−F(0)=0.60−0.35=0.25f(1)=F(1)-F(0)=0.60-0.35=0.25.

f(2)=F(2)−F(1)=0.85−0.60=0.25f(2)=F(2)-F(1)=0.85-0.60=0.25.

f(3)=F(3)−F(2)=1−0.85=0.15f(3)=F(3)-F(2)=1-0.85=0.15.

Step 2. Check. 0.15+0.20+0.25+0.25+0.15=1.000.15+0.20+0.25+0.25+0.15=1.00 ✓. …

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