Timeline for If $m(t)$ is the moment generating function of a random variable, then so is $(1/2)+(1/3)m(t)+(1/6)m(t)^2$. Explain why this is true
Current License: CC BY-SA 3.0
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| Oct 19, 2016 at 14:44 | comment | added | Glen_b | I think you made that slightly harder than it needs to be (though I guess it depends on your point of view) but something along the lines of what you were doing should work too. But do it more formally. | |
| Oct 19, 2016 at 14:40 | comment | added | tanner | So the mgf of Y could be m(t)= 1/2 + 1/2m(t), now if i look at my question I could say Y is a r.v with prob with 1/3, say X is a mgf with m(t), and I could define Z=X+Y so the mgf would look like mz(t)=1/2 + 1/3m(t) +(1/2)(1/3)mx+y(t) which i could show to be mz(t) = 1/2 +1/3m(t) + (1/6)m(t) ^ 2 . | |
| Oct 19, 2016 at 14:33 | comment | added | tanner | o ok i think im starting to get it so i defining mine in terms of random variables i could say that X is a r.v with mgf m(t), and Y =0 with probability 1/2. | |
| Oct 19, 2016 at 11:11 | history | edited | Glen_b | CC BY-SA 3.0 |
added 14 characters in body
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| Oct 19, 2016 at 9:03 | history | answered | Glen_b | CC BY-SA 3.0 |