4. Let Y be a positive-valued random variable, i.e., fy (y) = 0 for y a) €) yfr(y)dy > a f, yfr(y)dy = aP(Y > a) Thus P(Y α) €) = P((X – E[X])² > e?). Thus using the results of the previous question...

could you give me  a more detailed answer ?

Extracted 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?)