MHB Calculating Expectation of $X$ for a Nonnegative RV

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The discussion focuses on calculating the expectation of a nonnegative random variable (RV) defined as X=7·I_{X≤7}+7ε·I_{X>7}. Participants clarify the properties of expectation, specifically how to use indicator functions to determine probabilities. The correct answer to the posed exercise is identified as option (2), which relates the expectation of X to the probabilities of the events defined by the indicators. The conversation emphasizes understanding the calculations involved in transitioning from E(X) to E(1/ε·X). The final consensus confirms the correct application of expectation properties to arrive at the solution.
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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 $.
 
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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
 
Thanks, but I can not do the calculation because I have not understood the indicator operation.
 
If $A$ is an event, then $I_A(w) = 1$ if $w\in A$ and $0$ if $w\notin A$. So $E(I_A) = P(A)$.
 
Ok, thank you. In the case of exercise which calculations I have to do to get a reply? Excuse me, but I am stuck.
 
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.
 
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?
 
There are intermediate steps involved. What is $E(I_{X > 7})$? What is $E(I_{X \le 7})$? Answer those questions first.
 
Unfortunately I do not know to calculate it. Is $E(I_{X>7})=P(X>7)$ and $E(I_{X\leqslant 7})=P(X\leqslant 7)$?
 
  • #10
You are correct!
 
  • #11
And after?
 
  • #12
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

$$E(X) = 7P(X \leqslant 7) + 7\varepsilon P(X > 7)$$
 
  • #13
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?
 
  • #14
You correctly identified answer (2) as the correct answer choice. To get the that step, use property (2) of expectation.
 
  • #15
$E(tX)=tE(X)$
$E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }E(X)$
$E(X)=7P(X\leqslant 7)+7\varepsilon P(X>7)$
$E(\frac{1}{\varepsilon }X)=\frac{1}{\varepsilon }(7P(X\leqslant 7)+7\varepsilon P(X>7))$
$E(\frac{1}{\varepsilon }X)=\frac{7}{\varepsilon }P(X\leqslant 7)+7P(X>7)$
Correct?
 
  • #16
Yes, that's correct.
 
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