This is from the A/S/M study guide, and the answer is listed, I just don't understand how he's arriving at the answer... I'm sure it's something simple I am missing!
I have two identically distributed, independent random variables, and their Moment Generating Function is:
MGF of X+Y (t) = .09e^-2t+.24e^-t+.34+.24e^t+.09e^2t
Now, I recognize this as a uniform distribution, so I have
X+Y = { -2 with probability .09, -1 with probability .24, 0 with probability .34, 1 with probability .24, 2 with probability .09}
Now, the question asks to find the probability of X being less than or equal to 0.
Since X and Y are identically distributed, independent random variables, I know that the MGF of X+Y = (MGF of X)^2, meaning one way to solve this problem is to convert the MGF of X+Y into a square, however I'm a little scared to leave that as my only "method" for when I take exam p because it can get a bit tricky!
So, I have the following notes:
-2 = (-1,-1)
-1 = (0, -1), (-1,0)
0 = (0,0), (-1, 1), (1,-1)
1 = (0,1), (1,0)
2 = (1,1)
Thus, the possible values of X are -1, 0, 1. How do I find the probability of each value?!
I cannot seem to figure out how to go about it... for example, for X=-1 I would do the following:
(X=-1 when X+Y=-2) with probability 1
(X=-1 when X+Y=-1) with probability .5,
(X=-1 when X+Y=0) with probability .33333...
So, I would do (1)(.09)+(.5)(.24)+(.333...)(.34)=.32333...
However, the correct probability for x=-1 is .3
Thank you in advance!
Hint:
Call $p=P(X=1)=P(Y=1)$, $q=P(X=-1)=P(Y=-1)$, and $r=P(X=0)=P(Y=0)$ hence $r=1-p-q$ (why?).
Then $P(X+Y=2)=$ $____$ and $X+Y=2$ happens when $(X,Y)=$ $____$ hence $p=$ $____$. Likewise, $P(X+Y=-2)=$ $____$ and $X+Y=-2$ happens when $(X,Y)=$ $____$ hence $q=$ $____$. Deduce that $r=$ $____$.
Finally, answer the question.