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I have n items priced P1,P2...Pn. An item will get sold with a probability p. I need to find an expression for the expected value of the average selling price. Since, the items can be sold in 2^n ways,this seems very daunting.

My take All items are sold with probability p^n. Then i can pick item to remain unsold. This can happen in n ways. This will add n expressions to the EV expression.

Then 2 items can remain unsold. This will add nC2 items to the EV expression.And so on.

It expressions seems to become very big very quickly.

Is there something obvious i am missing?

kangkan Dc
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  • I'm voting to close this question as off-topic because it is about [math.se] instead of programming or software development. – Pang Jan 31 '18 at 06:27

1 Answers1

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I think you're overthinking the problem. let E[x] be the expected value of x.

(see https://en.wikipedia.org/wiki/Expected_value#Definition)

E[P1] = P1*p1 + 0*(1-p1) = P1*p1

Using the identity E[X+Y] = E[X] + E[Y] (means "the expected price of all items is the sum of expected price of each item"), you get

E[P1 + P2 + ... + Pn] = E[P1] + E[P2] + ... + E[Pn] = P1*p1 + P2*p2 + ... + Pn*pn

Or, if p is the same for all items,

E[P1 + P2 + ... + Pn] = (P1+P2+..+Pn)*p

tdurtsch
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  • I thought about that. But there seems to be a problem: If i have two items valued at $50 and $60,and they each will be sold with probability=0.1 .Now using the simpler logic, the Expected sale price comes to $11. But the Avg sale price is a random variable that takes value 50 with probability 0.09,60 with 0.09,55 with 0.01 and zero with 0.81. The expected value comes to $10.45. The differences becomes larger with more number of objects – kangkan Dc Jul 10 '15 at 07:58