Question

Let \( X \) and \( Y \) be two discrete random variables with the joint moment generating function \[ M_{X,Y}(t_1, t_2) = \left( \frac{1}{3} e^{t_1} + \frac{2}{3} \right)^2 \left( \frac{2}{3} e^{t_2} + \frac{1}{3} \right)^3, t_1, t_2 \in \mathbb{R}. \] Then \( P(2X + 3Y > 1) \) equals ......... 
 

Hint:

The moment generating function can be used to derive probabilities for sums and transformations of random variables.

Answer 1

Correct Answer: 0.97 - 0.99

Step 1: Identify the distribution from the MGF

The joint MGF is: $$M_{X,Y}(t_1, t_2) = \left(\frac{1}{3}e^{t_1} + \frac{2}{3}\right)^2 \left(\frac{2}{3}e^{t_2} + \frac{1}{3}\right)^3$$

This can be rewritten as: $$M_{X,Y}(t_1, t_2) = \left[\frac{1}{3}e^{t_1} + \frac{2}{3}e^{0}\right]^2 \left[\frac{2}{3}e^{t_2} + \frac{1}{3}e^{0}\right]^3$$

This is the MGF of independent random variables where:

• $X = X_1 + X_2$ where $X_1, X_2$ are i.i.d. Bernoulli with $P(X_i = 1) = \frac{1}{3}$, $P(X_i = 0) = \frac{2}{3}$
• $Y = Y_1 + Y_2 + Y_3$ where $Y_1, Y_2, Y_3$ are i.i.d. Bernoulli with $P(Y_j = 1) = \frac{2}{3}$, $P(Y_j = 0) = \frac{1}{3}$

Therefore:

• $X \sim \text{Binomial}(2, \frac{1}{3})$
• $Y \sim \text{Binomial}(3, \frac{2}{3})$
• $X$ and $Y$ are independent

Step 2: Find the distributions

For $X \sim \text{Binomial}(2, \frac{1}{3})$:

• $P(X = 0) = \binom{2}{0}\left(\frac{1}{3}\right)^0\left(\frac{2}{3}\right)^2 = \frac{4}{9}$
• $P(X = 1) = \binom{2}{1}\left(\frac{1}{3}\right)^1\left(\frac{2}{3}\right)^1 = \frac{4}{9}$
• $P(X = 2) = \binom{2}{2}\left(\frac{1}{3}\right)^2\left(\frac{2}{3}\right)^0 = \frac{1}{9}$

For $Y \sim \text{Binomial}(3, \frac{2}{3})$:

• $P(Y = 0) = \binom{3}{0}\left(\frac{2}{3}\right)^0\left(\frac{1}{3}\right)^3 = \frac{1}{27}$
• $P(Y = 1) = \binom{3}{1}\left(\frac{2}{3}\right)^1\left(\frac{1}{3}\right)^2 = \frac{6}{27} = \frac{2}{9}$
• $P(Y = 2) = \binom{3}{2}\left(\frac{2}{3}\right)^2\left(\frac{1}{3}\right)^1 = \frac{12}{27} = \frac{4}{9}$
• $P(Y = 3) = \binom{3}{3}\left(\frac{2}{3}\right)^3\left(\frac{1}{3}\right)^0 = \frac{8}{27}$

Step 3: Find $P(2X + 3Y > 1)$

We need $2X + 3Y > 1$, which is equivalent to $2X + 3Y \geq 2$.

The complement is $2X + 3Y \leq 1$:

• $2X + 3Y = 0$: requires $X = 0, Y = 0$
• $2X + 3Y = 1$: requires $X = 0, Y = 0$ (gives 0) or... no integer solution gives exactly 1

Actually, $2X + 3Y = 1$ requires $2X + 3Y = 1$. Since $X, Y$ are non-negative integers:

• If $Y = 0$: $2X = 1$ (no integer solution)
• If $Y = 1$: $2X = -2$ (impossible)

So $2X + 3Y = 1$ has no solutions.

Therefore: $P(2X + 3Y \leq 1) = P(2X + 3Y = 0) = P(X = 0, Y = 0)$

Since $X$ and $Y$ are independent: $$P(X = 0, Y = 0) = P(X = 0) \cdot P(Y = 0) = \frac{4}{9} \cdot \frac{1}{27} = \frac{4}{243}$$

Therefore: $$P(2X + 3Y > 1) = 1 - \frac{4}{243} = \frac{239}{243}$$

Step 4: Convert to decimal

$$\frac{239}{243} \approx 0.9835$$

Answer: $$P(2X + 3Y > 1) = \frac{239}{243} \approx 0.984$$