X is a geometric with p=1/3.
I've never seen this before.
The distribution of Y is interesting.
The hint is to go after the rv Y-X via its MGF.
I think you should try the conditional expectation
.
A fair die is rolled repeatedly and independently. Let X be the number of rolls until the first time a 5 or 6 has been observed. Note that the trial that 5 or 6 occurred is also counted. Let Y be the number of casts until the first time when both 5 or 6 have been observed.
a) What is the moment generating function of X, Mx(t)
b) What is the moment generating function of Y-X, My-x(t)
c) Find the moment generating function of Y, My(t) and use it to find E(Y)
Help would be greatly appreciated.
After doing part (a), you find a general form for the .
I did what utopiaNow did for part (b), and using the work from (a), I used p = 1/6
Part (c) is due to the fact that the geometric r.v. is memoryless.
Part (d), I multiplied to find the mgf of . To find the , I found the derivative of .
I'm not quite sure about my work for (d), but I think that's the gist of it... try it out
Thanks, nice catch. I tried taking the derivative of the mgf again, and got a final answer of 9.
I missed doing the chain rule while finding the derivative for a part of the question, but I fixed it. I just got caught up in the math, I forgot what the question was, and hence, didn't pick up that 9/64 was a weird answer...
Thanks for that! 9 does make more sense, and I fixed up my math.
K well for that we have to think about what X and Y are. X is a geometric r.v. with p1 = 1/3. Now Y is the total number of rolls till both 5 and 6 are observed. So for Y we keep rolling dice till we get a 5 or 6, ( which is X), then we keep rolling again till we get whichever one we didn't get on the first "success"( so we wait for 5 if we got 6 the first time and vice versa).
So now think about it like this:
---------->X------------>Y-X
-----------*-------------*
------------------------->Y
Now you can see that Y - X is just the difference, or the second wait period. And since the dice is being rolled independently, we can basically treat Y-X as rolling a dice until we get whichever 1 we didn't get the first time. Only 1/6 possibilities this time, since 5 or 6 already happens first at the first star(X). So p2 = 1/6.
Sorry..coz I am not familiar with how to use the math symbol...so i didnt post it...
My work:
let e^t = u du/dt = e^t
My(t) = 1/18 u^2 / ( 1-3/2 u+5/9 u^2)
Applying quotient rule:
--> (1/9u-1/12 u ^2) / (1-2/3u+5/9u^2)^2
Should I let t=0...then....it turns out... (1/9-1/12 )/(1-3/2+5/9)^2 = 9
am I on the right track?
Thanks!