Let Yn≥0.

a) show that E[Yn/(1+Yn)]->0 => Yn->0 in probability.

b) show that Yn->0 in probability => E[Yn/(1+Yn)]->0.

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a) Let ε>0.

Using Markov's inequality,

P(Yn≥ε) = P{[Yn/(1+Yn)] ≥ ε/(1+Yn)} ≤ E[Yn/(1+Yn)] /[ε/(1+Yn)]

If we can show that E[Yn/(1+Yn)] /[ε/(1+Yn)]->0, then this implies that Yn->0 in probability and we're done.

But how can we prove that E[Yn/(1+Yn)] /[ε/(1+Yn)]->0 ? I know that the numerator goes to 0 by assumption, but I am really worrying about the denominator; if Yn->∞ as n->∞, then the denominator would go to 0 as n->∞, and 0/0 is indeterminate!? How can we resolve this problem? I am really scared now, help...

b) For this part, it seems like the use of Markov's inequality wouldn't work because the inequality goes the "wrong" way, so I am now stuck and I don't know where to start. Could somebody please give me some hints?

Any help is very much appreciated!