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Let $(X,Y)$ be a random vector with joint moment generating function $$M(t_1,t_2) = \frac{1}{(1-(t_1+t_2))(1-t_2)}$$ Let $Z=X+Y$. Then, Var(Z) is equal to: (IIT JAM MS 2021, Q21)

Using $M_{X+Y}(t) = M_{X,Y}(t, t)$ gives: $$M(t,t) = \frac{1}{(1-2t)(1-t)}=\frac{1}{(1-2t)} \times \frac{1}{(1-t)}$$

And this can be split into two independent random variables with $$M_U(t)=\frac{1}{(1-2t)}=\frac{0.5}{(0.5-t)} \sim Exp(\lambda=0.5)$$ $$M_V(t)=\frac{1}{(1-t)} \sim Exp(\lambda=1)$$

$M_Z(t)=M_{U+V}(t)=M_U(t)M_V(t)$ which means $U$ and $V$ are independent, and hence the covariance terms is zero: $$V(Z)=V(U+V)=V(U)+V(V)+2Cov(U,V)=V(U)+V(V)=4+1=5$$

This matches the correct answer from the official answer key (which is indeed 5). However, I am skeptical about this solution since $X$ and $Y$ are independent, and this seems to be a way to convert it into two independent random variables. It gives the right answer, but is it valid? And if isn't till what part is it valid?

The general question of interest is that can two dependent random variables be split into two independent random variables using this kind of MGF manipulation.

Ѕᴀᴀᴅ
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Starlight
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1 Answers1

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I think your approach is fine. Here is a paraphrasing of your proof that hopefully makes it clearer:

  • You are given the MGF of $(X, Y)$. (They are dependent.)
  • You show that the MGF of $X+Y$ is $\frac{1}{(1-2t)(1-t)}$.
  • Separately, you show that if $U \sim \operatorname{Exponential}(0.5)$ and $V \sim \operatorname{Exponential}(1)$ are independent, then the MGF of $U+V$ is also $\frac{1}{(1-2t)(1-t)}$.
  • What remains is to show that "because $X+Y$ and $U+V$ have the same MGF, they must have the same distribution." This is the "uniqueness theorem" for MGFs; you can find some discussion and references to proofs here and here.

I guess this implies that decomposing an MGF into the product of two MGFs is a valid way to decompose a random variable into the sum of two independent random variables. One sometimes does this with joint PDFs as well: if a joint PDF can be decomposed into the product of marginal PDFs, then you can conclude the marginal random variables are independent, such as $U$ and $V$ here.

angryavian
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