# Calculate Probability from Moment Generating Function

Let 𝑋 and 𝑌 be two discrete random variables with the joint moment generating function

$$𝑀_{𝑋,𝑌}(t_{1},t_{2})=(\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2} (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3} ,t_{1},t_{2} \in R.$$ Then $$P$$ $$( 2𝑋 + 3𝑌 > 1)$$ equals

My Approch:

$$𝑀_{𝑋} = 𝑀_{𝑋,𝑌}(t_{1},0) = (\frac{1}{3} e^{t_{1}} + \frac{2}{3})^{2}$$ similar to MGF of $$Binomial(2,\frac{1}{3})$$

$$𝑀_{Y} = 𝑀_{𝑋,𝑌}(0,t_{2}) = (\frac{2}{3} e^{t_{2}} + \frac{1}{3})^{3}$$ similar to MGF of $$Binomial(3,\frac{2}{3})$$

$$P$$ $$( 2𝑋 + 3𝑌 > 1) = 1 - P(X=0) P(Y=0) = 1 - (\frac{2}{3})^{2}(\frac{1}{3})^{3} \simeq 0.98$$

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?

• Why is $n=2$ at $Y$? At the mgf of $Y$ the exponent is $2$, a typo? – callculus Feb 5 at 17:20
• Thanks for pointing out. Corrected the same. – Pradeep Bihani Feb 5 at 17:24
• 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. – Robert Israel Feb 5 at 17:48
• The rest looks correct to me. I cannot think of a better approach. – callculus Feb 5 at 17:50