Hello,

Let

Then we want to show that in probability iff

Note that

So f is a strictly increasing bijection from to

So we have (1)

And similarly, we have (2)

a) You indeed have to use Markov's inequality, but I think you did it strangely... because when given P(|X|>a), a should be a constant and not a random variable, like in what you did.

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From (1), we have , which, by Markov's inequality, is

But we supposed that . Thus

And this means that in probability.

b) This one is more tricky imo...

- We can see that

- Note that almost surely (this follows from the property of the function f)

Hence

But by (2),

So finally, we get

Using the fact that in probability,

And we're left with

Thus