Calculating Expectation of $X$ for a Nonnegative RV

[Francobati]

Good morning. Can you help me to solve this exercise. The correct answer should be the 2, but how is it calculated? Thanks.

Let $l_{+}$ be the set of nonnegative simple rv’s. Pick $X=7\cdot I _{\left \{ X\leqslant 7 \right \}}+7\varepsilon \cdot I_{\left \{ X> 7 \right \}}\epsilon l_{+}$ , for $\varepsilon > 0$. What statement is TRUE?

(1): $E(X)=P(X>7)$;
(2): $E(\frac{1}{\varepsilon }{X})=\frac{7}{\varepsilon }P(X\leqslant 7)+7P(X> 7)$;
(3): $E(X)=0$;
(4): $\frac{E(X)}{7}=P(X\geqslant 7)$, provided that $\varepsilon \rightarrow 0$;
(5): $E(X)=\varepsilon $.

 

[Euge]

Hi Francobati,

Welcome. (Wave) Your problem can be solved using the following properties of expectation.

1. $E(X + Y) = E(X) + E(Y)$

2. $E(tX) = tE(X)$ ($t$ is a constant)

Give it a try, and if you have any questions then let me know. :D

 

[Francobati]

Thanks, but I can not do the calculation because I have not understood the indicator operation.

 

[Euge]

If $A$ is an event, then $I_A(w) = 1$ if $w\in A$ and $0$ if $w\notin A$. So $E(I_A) = P(A)$.

 

[Francobati]

Ok, thank you. In the case of exercise which calculations I have to do to get a reply? Excuse me, but I am stuck.

 

[Euge]

Using the addition property (property 1) for expectation, we have

$$E(X) = E(7I_{\{X \le 7\}}) + E(7\varepsilon I_{\{X > 7\}})$$

By property 2 of expectation,

$$E(7I_{\{X \le 7\}}) = 7E(I_{\{X\le 7\}})\quad \text{and}\quad E(7\varepsilon I_{\{X > 7\}}) = 7\varepsilon E(I_{\{X > 7\}}).$$

Thus

$$E(X) = 7E(I_{\{X \le 7\}}) + 7\varepsilon E(I_{\{X > 7\}}).$$

Now use the result $E(I_A) = P(A)$ for any event $A$ to determine which answer choice is best.

 

[Francobati]

Thanks. As a step from $E(X)=7E(I_{{X\leqslant 7}})+7\varepsilon E(I_{X> 7})$ to $E(\frac{1}{\varepsilon }{X})=\frac{7}{\varepsilon }P(X\leqslant 7)+7P(X> 7)$;? I do some recollection to common factor? And what?

 

[Euge]

There are intermediate steps involved. What is $E(I_{X > 7})$? What is $E(I_{X \le 7})$? Answer those questions first.

 

[Francobati]

Unfortunately I do not know to calculate it. Is $E(I_{X>7})=P(X>7)$ and $E(I_{X\leqslant 7})=P(X\leqslant 7)$?

 

[Euge]

You are correct!

 

[Francobati]

And after?

 

[Euge]

Well, look at the last equation in Post #6. Since $E(I_{X \leqslant 7}) = P(X \leqslant 7)$ and $E(I_{X > 7}) = P(X > 7)$, then

$$E(X) = 7P(X \leqslant 7) + 7\varepsilon P(X > 7)$$

 

[Francobati]

Ok, thanks. Perfect! Now I have to go by $E(X)=7P(X\leqslant 7)+7\varepsilon P(X>7)$ to $E(\frac{1}{\varepsilon}X)=\frac{7}{\varepsilon}P(X\leqslant 7)+7P(X>7)$. What calculations do I need to do? There is perhaps some recollection to common factor to do?

 

[Euge]

You correctly identified answer (2) as the correct answer choice. To get the that step, use property (2) of expectation.

 

[Francobati]

$E(tX)=tE(X)$
$E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }E(X)$
$E(X)=7P(X\leqslant 7)+7\varepsilon P(X>7)$
$E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }(7P(X\leqslant 7)+7\varepsilon P(X>7))$
$E(\frac{1}{\varepsilon }X)=\frac{7}{\varepsilon }P(X\leqslant 7)+7P(X>7)$
Correct?

 

[Euge]

Yes, that's correct.