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Let ๐‘‹ and ๐‘Œ be two discrete random variables with the joint moment generating function

$$๐‘€_{๐‘‹,๐‘Œ}(t_{1},t_{2})=(\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2} (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3} ,t_{1},t_{2} \in R.$$ Then $P$ $( 2๐‘‹ + 3๐‘Œ > 1)$ equals

My Approch:

$ ๐‘€_{๐‘‹} = ๐‘€_{๐‘‹,๐‘Œ}(t_{1},0) = (\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2} $ similar to MGF of $Binomial(2,\frac{1}{3})$

$ ๐‘€_{Y} = ๐‘€_{๐‘‹,๐‘Œ}(0,t_{2}) = (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3} $ similar to MGF of $Binomial(3,\frac{2}{3})$

$P$ $( 2๐‘‹ + 3๐‘Œ > 1) = 1 - P(X=0) P(Y=0) = 1 - (\frac{2}{3})^{2}(\frac{1}{3})^{3} \simeq 0.98 $

Is this a right approach to solve for this kind type of question?

What if in case, I don't know the MGF of known distribution?

Is there any general way to solve this kind of problems?

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  • $\begingroup$ Why is $n=2$ at $Y$? At the mgf of $Y$ the exponent is $2$, a typo? $\endgroup$ – callculus Feb 5 at 17:20
  • $\begingroup$ Thanks for pointing out. Corrected the same. $\endgroup$ – Pradeep Bihani Feb 5 at 17:24
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    $\begingroup$ Since the joint MGF factors as the product of a function of $t_1$ and a function of $t_2$, each of which is an MGF, your random variables are independent. $\endgroup$ – Robert Israel Feb 5 at 17:48
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    $\begingroup$ The rest looks correct to me. I cannot think of a better approach. $\endgroup$ – callculus Feb 5 at 17:50

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