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Adding two distributions with same moment generating function

  • Thread starter trojansc82
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  • #1
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Homework Statement



I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

Mx(t) + My(t) if Mx(t) = (1/3 + 2/3et) and My(t) = (1/3 + 2/3et)

Homework Equations





The Attempt at a Solution



Would the solution be (1/3 + 2/3et)2?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement



I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

Mx(t) + My(t) if Mx(t) = (1/3 + 2/3et) and My(t) = (1/3 + 2/3et)

Homework Equations




The Attempt at a Solution



Would the solution be (1/3 + 2/3et)2?
No. Denoting the mgf of X by [itex]M_X(t)[/itex] we have the relationship
[tex]M_{a+bX}(t) = E(e^{t(a+bX)}) = E(e^{at}e^{btX})=e^{at}E(e^{btX})
=e^{at}M_X(bt)[/tex]

You are asking about the mgf of X + X = 2X. So using the above with a = 0 and b = 2 gives [itex]M_{2X}(t) = M_X(2t)[/itex].
 
  • #3
Ray Vickson
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Homework Statement



I wanted to know what the result would be if you added two distributions with the same moment generating function.

For example, what would the result be of:

Mx(t) + My(t) if Mx(t) = (1/3 + 2/3et) and My(t) = (1/3 + 2/3et)

Homework Equations





The Attempt at a Solution



Would the solution be (1/3 + 2/3et)2?
There is a difference between adding distributions (or mgf's) and adding random variables. Which do you mean? If you want to find the MGF of X+Y, and assuming X,Y are independent, you get (1/3 + 2/3*exp(t))^2. If you want to find the MGF of the sum of the distributions of X and Y, you get 2*(1/3 + 2/3*exp(t)), but this thing does not really have much meaning.

RGV
 

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