Expected value prove problem

Question

So the question asks: Let Y ≥ 0 be a non-negative random variable. Prove that that for any $t > 0$,

                              P (Y ≥ t) ≤ E [Y ]/t

So so far I have:

Define the event $E ={Y≥t}$, and Indicator $EI_E=P(Y≥t)$, To show $ EI_E[Y]≤ E [Y ]/t$

So I want to show since

$ I_Aw = 1$ when $w ∈ A$

$ I_Aw = 0 $ when $w ∈ A^c$

So $I_E≤ 1$

$Y ≥ Y · I_E$

And since $Y≥t$,

$Y · I_E≥ t · I_E$

so $Y ≥ Y · I_E≥ t · I_E$

So $Y ≥ t · I_E$

And $E[Y] ≥ E[t] · E[I_E]$

So $ E[Y] ≥ t· P(Y≥t)$

Which proved that $ P (Y ≥ t) ≤ E [Y ]/t $

But I am not sure if my solution is right, and especially the part that I put E the expected value to both sides of the equations. Is it OK to do so?

Answer 1

I think your proof works, apart from three points:

• Using $E$ both for an event and for expectation may not be the best choice

• your phrase "And since $Y≥t$" might be better said as "And since $Y≥t$ when $I_E\not = 0$"

• you might need to justify asserting that an expectation does not alter the direction of an inequality

Wikipedia uses a similar approach to the Markov inequality, and some others