# Thread: Show Yn->0 in probability iff E[Yn/(1+Yn)]->0

1. ## Show Yn->0 in probability iff E[Yn/(1+Yn)]->0

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!

2. Hello,

Let $\displaystyle Z_n=\frac{Y_n}{1+Y_n}=f(Y_n)$

Then we want to show that $\displaystyle Y_n\to 0$ in probability iff $\displaystyle E[Z_n]\to 0$

Note that $\displaystyle f'(y)=\frac{1}{(1+y)^2}>0$

So f is a strictly increasing bijection from $\displaystyle [0,\infty)$ to $\displaystyle [0,1)$

So we have $\displaystyle \{Y_n>\epsilon\}=\{f(Y_n)>f(\epsilon)\}=\left\{Z_n >\tfrac{\epsilon}{1+\epsilon}\right\}$ (1)

And similarly, we have $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ (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 $\displaystyle \frac{\epsilon}{1+Y_n}$ in what you did.

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From (1), we have $\displaystyle P(|Y_n|>\epsilon)=P(Y_n>\epsilon)=P\left(Z_n>\tfra c{\epsilon}{1+\epsilon}\right)$, which, by Markov's inequality, is $\displaystyle \leq \frac{1+\epsilon}{\epsilon} \cdot E[Z_n]$

But we supposed that $\displaystyle E[Z_n]\to 0$. Thus $\displaystyle P(|Y_n|>\epsilon)\to 0~,~ \forall \epsilon>0$

And this means that $\displaystyle Y_n\to 0$ in probability.

b) This one is more tricky imo...

$\displaystyle E[Z_n]=E[Z_n\cdot\bold{1}_{Z_n\leq \epsilon}]+E[Z_n\cdot\bold{1}_{Z_n>\epsilon}]$

- We can see that $\displaystyle E[Z_n\cdot\bold{1}_{Z_n\leq \epsilon}]\leq E[\epsilon \cdot \bold{1}_{Z_n\leq \epsilon}]=\epsilon \cdot P(Z_n\leq \epsilon)$

- Note that $\displaystyle Z_n<1$ almost surely (this follows from the property of the function f)
Hence $\displaystyle E[Z_n\cdot\bold{1}_{Z_n>\epsilon}]<P(Z_n>\epsilon)$
But by (2), $\displaystyle P(Z_n>\epsilon)=P\left(Y_n>\tfrac{\epsilon}{1-\epsilon}\right)=P(Y_n>\epsilon')$

So finally, we get $\displaystyle E[Z_n]\leq \epsilon+P(Y_n>\epsilon')$

Using the fact that $\displaystyle Y_n\to 0$ in probability, $\displaystyle P(Y_n>\epsilon')\to 0$

And we're left with $\displaystyle 0\leq \lim_{n\to\infty} E[Z_n]\leq \epsilon~,~ \forall \epsilon>0$

Thus $\displaystyle \lim_{n\to\infty} E[Z_n]=0$

3. Thank you, Moo! It's actually a lot trickier than I originally thought...

I have a question about (2): $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$.

Zn>ε
<=> Yn/(1+Yn)>ε
<=> Yn > ε + εYn
<=> Yn(1-ε) > ε
<=> Yn > ε/(1-ε) only when ε<1
What if ε>1??? Then it would be Yn < ε/(1-ε) ???

So $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ is not true for all ε>0?

4. Originally Posted by kingwinner
Thank you, Moo! It's actually a lot trickier than I originally thought...

I have a question about (2): $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$.

Zn>ε
<=> Yn/(1+Yn)>ε
<=> Yn > ε + εYn
<=> Yn(1-ε) > ε
<=> Yn > ε/(1-ε) only when ε<1
What if ε>1??? Then it would be Yn < ε/(1-ε) ???

So $\displaystyle \{Z_n>\epsilon\}=\left\{Y_n>\tfrac{\epsilon}{1-\epsilon}\right\}$ is not true for all ε>0?
Good remark

Actually, in general, we need "small" epsilons, as we're interested in what happens when epsilon gets very very small.
So you can consider $\displaystyle \epsilon <1$ without much loss of generality ^^