problem 15

6.2.1 problem 15

(a). Since the MGF is uniquely determined from the distribution, and X and Y are identically distributed,

M_{Y^{2}}^{2} (t) = M_{X^{2}}^{2} (t) = {(1 - 2 t)}^{- \frac{1}{2}} .

As X and Y are mutually independent,

M_{W} (t) = M_{X^{2} + Y^{2}}^{2} (t) = M_{X^{2}}^{2} (t) \cdot M_{Y^{2}}^{2} (t) .

Hence,

M_{W} (t) = \frac{1}{(1 - 2 t)} .

(b). Exponential distribution, with $λ = \frac{1}{2}$ . Note that this is a special case (as mentioned in the question) of the Gamma distribution (Chapter 8).

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