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Distribution with pmf and rand. variables. |
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| Oct12-10, 08:34 PM | #1 |
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Distribution with pmf and rand. variables.
I posted this in the wrong section before and meant to put it here, so i apologize if you seen this before.
X=demand for the magazine with pmf x | 1 2 3 4 p(x)| .1 .2 .4 .3 Shop owner pays $1.00 for each copy of mag. and charges $2.00. If mags. left at end of week are not worth anything, is it better to order two, three, or four copies of the mag.? I know i need to introduce the random variables: Y_k = # of mags. sold R_k= the net profit if k mags are ordered. I am NOT trying to just get the answer out of someone, I just need advice on how to start this.. Do I need to make another pmf for Y_k and R_k? Or do I need to figure out expected value. just a hint may help me understand this problem |
| Oct12-10, 09:31 PM | #2 |
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It looks like an open-ended question. First step could be to write down the profit for all 16 combinations {(1 bought, 1 sold), (1 bought, 2 sold), ...} perhaps as a 4x4 table.
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| Oct12-10, 09:38 PM | #3 |
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i dont think its open ended, because R_k= -1k+2*Y_k since R_k is the profit
could i simply find the expected value is 1, 2, or 3 are sold thats it? |
| Oct12-10, 10:23 PM | #4 |
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Distribution with pmf and rand. variables.
That's the open-ended part, it's up to you to choose a selection criteria. Expected value is only one of infinitely many possibilities. It's good that you've got a formula for the profit though it's important to actually look at the values and their relative probabilities (for example, with an appropriate chart) otherwise important details can be hidden.
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| Oct12-10, 10:30 PM | #5 |
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okay the profit for k=2 i got 3.8
when i am calculating it for when k=3 is this equation correct? -1(3)+2(.1*1+.2*3.8+.4*3.8+.3*3.8) i may be going off a longshot but i used the profit from k=2 for the values of x in this equation. I just want to make sure im doing k=3 right so i can figure out when k=4... |
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