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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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Concept understanding — Probability Density Function & Cumulative Distribution

Cumulative distribution function (both cases). For any random variable XX, the cdf F(x)=P(X≤x)F(x)=P(X\le x) is defined for every real xx.

  • Discrete (Definition 11.4): F(x)=∑xi≤xf(xi)F(x)=\sum_{x_i\le x}f(x_i) — a step function, constant between support points and jumping by f(xi)f(x_i) at each xix_i. Conversion both ways: given the pmf, FF is the running (cumulative) sum of ff up to xx; given FF, the pmf is recovered as the jump size f(xi)=F(xi)−F(xi−1)f(x_i)=F(x_i)-F(x_{i-1}) at each point of discontinuity (with F(x0)=0F(x_0)=0 before the first jump) — the jump of FF at aa is exactly P(X=a)P(X=a).
  • Continuous (Definition 11.7): F(x)=∫−∞xf(u) duF(x)=\int_{-\infty}^x f(u)\,du — here FF is everywhere continuous (no jumps, since no single point carries probability). Conversion both ways: given the pdf, integrate piece by piece to build FF; given FF, differentiate — f(x)=F′(x)f(x)=F'(x) wherever the derivative exists (at the finitely many "corner" points, ff may be set to any convenient value, since it never affects an interval probability). …

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