I have scoured my textbook for the concept alluded to in the title of this thread; however, my textbook has failed, in that it provides no such information.

Does anyone know of some resources for this concept?


For instance, let me post a problem I am working on:

An instructor has given a short test consisting of two parts. For a randomly selected student, let $X=$ the number of points earned on the first part and $Y=$ the number of points earned on the second part. Suppose that the joint pmf of X and Y is given in the accompanying table.

p(x,y) 0 | 5 | 10 | 15

0 | 0.02| 0.06| 0.02| 0.10

5 | 0.04| 0.15| 0.20| 0.10

10| 0.01| 0.15| 0.14| 0.01

a) If the score recorded in the grade book is the total number of points earned on two parts, what is the distribution of $X+Y$ and what is the expected recorded score $E(X+Y)$

b) If the maximum of the two scored is recorded, what is the distribution of the maximum score $Max(X,Y)$ ad the expected maximum score $E(Max(X,Y))$

Now that I re-look at this question, I doubt whether I can solve any of it. At any rate, how do I find the distribution of $X+Y$?

  • $\begingroup$ more details would be needed to answer or provide help else I am afraid this post will get closed. $\endgroup$ – jay-sun Apr 20 '13 at 22:05
  • $\begingroup$ Thank you for the suggest, @jay-sun. I will be edited momentarily. $\endgroup$ – Mack Apr 20 '13 at 22:36

Hint for a) $E(X+Y)=\sum_{x,y} (x+y)p(x,y)$ You can substitute for every value of $x$ and $y$ and you know $p(x,y)$.

Hint for b) Same as above, calculate $E(max(x,y))=\sum_{x,y} max(x,y)p(x,y)$

  • $\begingroup$ Thank you, that certainly helps for the most part; but how do I find the probability distribution $X+Y$? $\endgroup$ – Mack Apr 21 '13 at 12:27

The distribution of X+Y is formally given by discrete convolution; practically this is the sum of each case contributing to the given sum. P (X+Y=z) = SUM p(x,y) where x+y = z e.g. here P(X+Y=5) = p(0,5) + p(5,0) = 0.04 + 0.06 = 0.10 P(X+Y=10) = p(0,10) + p(5,5) + p(10,0) etc


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