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#1
11-20-2017, 01:18 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9
expected value

A company agrees to accept the highest of four sealed bids on a property. The four bids are regarded as four independent random variables with common cumulative distribution function
F(x) =
1/2
(1 + sinπx),
3/2 ≤ x ≤ 5/2 and 0 otherwise. What is the expected value of the accepted bid?

My question is how can the four bids be different if they have the same cdf?
#2
11-20-2017, 01:27 PM
 Gandalf Site Supporter Site Supporter SOA Join Date: Nov 2001 Location: Middle Earth Posts: 30,793

The same cdf means that (for example) the probability that #1 is less than 2 equals the probability that #2 is less than 2. Compare to flipping 2 fair coins 1 time each and counting the number of heads of each. Each has F(0)=.5, but the results of the actual flips might be different for the two coins.
#3
11-20-2017, 01:52 PM
 Vorian Atreides Wiki/Note Contributor CAS Join Date: Apr 2005 Location: Hitler's Secret Bunker Studying for ACAS College: Hard Knocks Favorite beer: Sam Adams Cherry Wheat Posts: 60,645

What is the probability that bid #1 is less than 1.6?

What is the probability that bid #2 is at least 1.6 but less than 1.75?

What is the probability that bid #3 is greater than 1.8 but no more than 2.0?

What is the probability that bid #4 is greater than 2.0?
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#4
11-20-2017, 02:38 PM
 Marcie Member CAS Join Date: Feb 2015 Posts: 6,621

Quote:
 Originally Posted by efast A company agrees to accept the highest of four sealed bids on a property. The four bids are regarded as four independent random variables with common cumulative distribution function F(x) = 1/2 (1 + sinπx), 3/2 ≤ x ≤ 5/2 and 0 otherwise. What is the expected value of the accepted bid? My question is how can the four bids be different if they have the same cdf?
Reminds me of a story...

I was playing poker with three of my buddies the other night, and we just kept getting dealt the exact same cards! Every time - all four of us!

The other three thought that was strange, but I, being the actuary at the table, calmly explained: it's because we all have the same exact probability distribution, so why wouldn't we all get the exact same cards?
#5
11-20-2017, 03:45 PM
 efast Non-Actuary Join Date: Nov 2017 Posts: 9

#6
11-20-2017, 04:03 PM
 Z3ta Member SOA Join Date: Sep 2015 Posts: 361

Consider the number of heads flips in the flip of a single coin (either 1 or 0).

The PMF is
$
P(X=x)=\begin{cases}
\frac{1}{2}, &x=0,1\\
0, &\text{otherwise}
\end{cases}$

The number of tails flips in the same flip has this PMF as well.

Both of these random variables have the same PMF (and CDF), but they are necessarily never equal to one another because if the coin lands on heads, then it didn’t land on tails.

My example is two random variables that are completely dependent on one another. In your problem they are specified to be independent. They are independent and identically distributed so they have the same probabilities for values they may realize, but the outcomes of any do not inform the outcomes of the others.

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