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Math Help - Distribution with pmf and cdf

  1. #1
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    Distribution with pmf and cdf

    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 variable Y_k = # of mags. sold, while R_k= the net profit if k mags are ordered.

    So do I need to find the probability distributions for k=2,3,4,5 in order to answer the quesiton?? I am just lost on how to start this or how to get the pmf for Y
    Last edited by mr fantastic; October 11th 2010 at 05:58 PM.
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  2. #2
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    This problem has no continous properties, thus you have to calculate the expected profit for every possible number of magazins ordered. The shop owner will choose the option that yields the highest profits. The expected payoff is the expected profit (which is bounded in every case by the number of magazins ordered. Thus e.g. the exp. profit for 2 ordered magazines will be: (0.1 * 1 + 0.2 * 2 + 0.4 * 2 + 0.3 * 2) * 2 - 1* 2. Btw: For a more general case the notation would be: (0.1 * 1 + 0.2 * min(ordered,2) + 0.4 * min(ordered,3) + 0.3 * min(ordered,4) * 2 - 1 * ordered. The shop owner will choose this number of magazines that yields the highest profit, this variable is fixed since the shop owner only knows the distribution of demand.

    You cannot however introduce a variabiable Y_k like you claimed since this is not a random event anymore but something you have to find a decision for.
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  3. #3
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    well our professor gave us a "hint" and said we need to introduce the random variable Y_k and R_k
    can i not make a pmf for Y_k and R_k??
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  4. #4
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    Well, you of course can define specific distribution and probability function, but I would not consider this very helpful for solving the problem.

    For Y_k however, I implicitly defined a probability function by limiting the probability to the constraint of sold magazines:

    A probability function would look like this: The probability that 1 magazine is sold is 100% (.1 + .2 + .4 + .3) (certain event), for 2 magazines the prob is: (.2 + .4 + .3) and so forth.

    For R_k you can also express the probability for a profit to be
    2 - ordered with prob 0.1
    2* min(ordered,2) - ordered by 0.2
    2*min(ordered,3) - ordered by 0.4
    2* min(ordered,4) - ordered by 0.3

    where you have to maintain "ordered" as a yet unknown variable. If you want to dertmine what option the shopkeeper will choose, refere to what I wrote before.
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  5. #5
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    so from that i need to calculate R_k for each scenario and determine which comes back with the most expected profit?
    i feel like i should use expected value for distributions
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  6. #6
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    If you want to calculate the expected profit, just add up all R_k with their probability. Then you have the expected proit dependend on the variable "ordered". Fill in 1-4 to see what maximizes your profit by comparison.
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  7. #7
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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...

    but going off what i did i got that they should order 3 copies..
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  8. #8
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    You are right about the profit for k=2. The formula you created for k=3, however, is not correct. Moreover, it does not make any sence. You can construct the profits in a sequence but I would rather just calculate the profit. If you want to do this, however, make sure to think about the change in profits. (a) What do I by a certain probability earn more with an additional magazine and (b) what do I pay more (for certain). Then add (a) and substract (b) to what you had for k=2.
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  9. #9
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    so for k=3 i dont use the formula -1(3) +2(.1*1+.2*2+.4*3+.3*4) like for k=2?

    hmm so maybe k=3 would be -1(3)+2(.1*2+.2(3)+.4(4)+.3(5)) ? i added one to the x value... im out of ideas after that one
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  10. #10
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    yes you do. but you do not use -1(3)+2(.1*1+.2*3.8+.4*3.8+.3*3.8) what differs from -1(3) +2(.1*1+.2*2+.4*3+.3*4) which is not correct either. The reason for his: Even if you could sell 4 magazines (what happens with the probability of .3 you cannot sell 4 if you ordered only 3. Thus it must be: -1(3) +2(.1*1+.2*2+.4*3+.3*3) for k = 3 or -1(3) +2(.1*1+.2*2+.4*2+.3*2). The formula you just stated was correct however for k = 4. The reason always is that you cannot sell more magazines than you ordered. This binds your maximall sells.
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  11. #11
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    for k=3
    -1(3) +2(.1*1+.2*2+.4*2+.3*2)

    for k=4
    -1(4) +2(.1*1+.2*2+.4*3+.3*3)

    correct? if so then this is starting to make A LOT more sense
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  12. #12
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    actually for the values after .2 .4 and .3 should those be 3 when k=3?
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  13. #13
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    almost there. it is like that:

    for k=3
    -1(3) +2(.1*1+.2*2+.4*3+.3*3)
    meaning: You buy 3 magazines and pay 1 dollar each, therefore you "earn" a negative -1 dollar * (3 magazines).
    then with probability of .1 one customer comes to your store. you can sell one of your three magazines. you earn by probability 0.1 * 1 magazines * 2 dollars
    with probability .2 two customers come to your store and you sell 2 of your three mags. you earn by this probability 0.2 * 2 magazines * 2 dollars
    with probability .4 three customers come to your store and you sell all 3 magazines. you earn by this probability 0.4 * 3 magazines * 2 dollars
    with probability .3 four customers come to your store and you could sell 4 magazines, but you only got 3 of them, so you still only sell 3 magazines since you have to send one of them home empty handed. so by probability 0.3 * 3 magazines * 2 dollars.

    since the 2 dollars are always the same we got them outside the brackets.

    for k=4
    -1(4) +2(.1*1+.2*2+.4*3+.3*4)
    meaning: You buy 4 magazines and pay 1 dollar each, therefore you "earn" a negative -1 dollar * (4 magazines).
    then with probability of .1 one customer comes to your store. you can sell one of your four magazines. you earn by probability 0.1 * 1 magazines * 2 dollars
    with probability .2 two customers come to your store and you sell 2 of your four mags. you earn by this probability 0.2 * 2 magazines * 2 dollars
    with probability .4 three customers come to your store and you sell 3 of your four magazines. you earn by this probability 0.4 * 3 magazines * 2 dollars
    with probability .3 four customers come to your store and you could sell 4 magazines and now you can. so by probability 0.3 * 4 magazines * 2 dollars.

    again we get the 2 dollars outside the bracket.
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  14. #14
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    wow, i get it!! thank you so much!
    so its actually better for the store owner to order 2 copies overall
    k=2 profit=3.8
    k=3 profit=2.2
    k=4 profit=1.8
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  15. #15
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    It is, you got it.
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