4. Let Y be a positive-valued random variable, 1.e., Jy E(Y) 4.1 Let a be any positive constant, and show that P(Y > a) < (Markov a inequality) 4.2 Let X be any random variable with variance o?, define Y =(X-E[X])2 and a = e2 for some ɛ. Obviously, the conditions of the problem are satisfied for Y and a as chosen here Derive the Chebychev inequality e

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4. Let Y be a positive-valued random variable, i.e., fy (y) = 0 for y < 0
E(Y)
4.1 Let a be any positive constant, and show that P(Y > a) <22 (Markov
a
inequality) -
4.2 Let X be any random variable with variance o?, define Y =(X-E[X])2 and a =
e? for some ɛ. Obviously, the conditions of the problem are satisfied for Y and a as
chosen here Derive the Chebychev inequality
P(|X – E(X)| > €) <
e2
Answer:
E[Y] = ° yfr(y)dy > yfr(y)dy
> a f, yfr(y)dy = aP(Y > a)
Thus P(Y α) < E[Υ]/α.
2) Clearly P(|X – E[X]| > €) = P((X – E[X])² > e?). Thus using the results of the previous question
we obtain
E[(X – E[X])²]
E2
P(|X – E[X]| > €) = P ((X – E[X])² > e?) <
Transcribed Image Text:4. Let Y be a positive-valued random variable, i.e., fy (y) = 0 for y < 0 E(Y) 4.1 Let a be any positive constant, and show that P(Y > a) <22 (Markov a inequality) - 4.2 Let X be any random variable with variance o?, define Y =(X-E[X])2 and a = e? for some ɛ. Obviously, the conditions of the problem are satisfied for Y and a as chosen here Derive the Chebychev inequality P(|X – E(X)| > €) < e2 Answer: E[Y] = ° yfr(y)dy > yfr(y)dy > a f, yfr(y)dy = aP(Y > a) Thus P(Y α) < E[Υ]/α. 2) Clearly P(|X – E[X]| > €) = P((X – E[X])² > e?). Thus using the results of the previous question we obtain E[(X – E[X])²] E2 P(|X – E[X]| > €) = P ((X – E[X])² > e?) <
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