Question

Let (X, Y) be a random vector having the joint moment generating function
\(M(t_1,t_2)=(\frac{1}{2}e^{-t_1}+\frac{1}{2}e^{t_1})^2(\frac{1}{2}+\frac{1}{2}et_2)^2,\ \ (t_1,t_2)\in \R^2.\)
Then P( |X + Y| = 2) equals __________ (round off to 2 decimal places)

Answer 1

Correct Answer: 0.23

To solve for \( P(|X+Y|=2) \) using the joint moment generating function (MGF) of \((X, Y)\), we start by identifying the distributions of \(X\) and \(Y\) from the given MGF. The MGF is:
\(M(t_1,t_2)=(\frac{1}{2}e^{-t_1}+\frac{1}{2}e^{t_1})^2(\frac{1}{2}+\frac{1}{2}e^{t_2})^2.\)

1. The form \(\frac{1}{2}e^{-t_1}+\frac{1}{2}e^{t_1}\) suggests a Bernoulli variable where \(X\) can be \(-1\) or \(1\) with probability \(\frac{1}{2}\).

2. Similarly, \((\frac{1}{2}+\frac{1}{2}e^{t_2})\) suggests a Bernoulli variable where \(Y\) can be \(0\) or \(1\) with probability \(\frac{1}{2}\).

Therefore, possible values for \((X,Y)\) are \((-1,0), (1,0), (-1,1), (1,1)\).

Next, compute \(X+Y\) for each case:

(X, Y)	X+Y
(-1, 0)	-1
(1, 0)	1
(-1, 1)	0
(1, 1)	2

We seek \(P(|X+Y|=2)\). This occurs when \(X+Y = -2\) or \(X+Y = 2\).

From our values of \(X+Y\): only \( (1, 1) \) yields \(X+Y=2\).

Thus, \(P(X+Y=2) = P((1, 1)) = \frac{1}{2} \times \frac{1}{2} = \frac{1}{4} = 0.25.\)

Hence, \(P(|X+Y|=2) = 0.25\).

Final Answer: \(P(|X+Y|=2) = 0.25\)