Q5 p290
Consider the following two experiments: the first has outcome X
taking on the values 0, 1, and 2 with equal probabilities; the second
results in an (independent) outcome Y taking on the value 3 with
probability 1/4 and 4 with probability 3/4. Find the distribution of
- Y + X
Y + X could be the values {3,4,5,6}
For P(Y + X = 3): P(X = 0)P(Y = 3) = \(\frac{1}{3}*\frac{1}{4}\) = \(\frac{1}{12}\)
For P(Y + X = 4): P(X = 0)P(Y = 4) + P(X = 1)P(Y = 3) = \(\frac{1}{3}*\frac{3}{4}\) + \(\frac{1}{3}*\frac{1}{4}\) = \(\frac{3}{12}\) + \(\frac{1}{12}\) = \(\frac{1}{3}\)
For P(Y + X = 5): P(X = 1)P(Y = 4) + P(X = 2)P(Y = 3) = \(\frac{1}{3}*\frac{3}{4}\) + \(\frac{1}{3}*\frac{1}{4}\) = \(\frac{3}{12}\) + \(\frac{1}{12}\) = \(\frac{1}{3}\)
For P(Y + X = 6): P(X = 2)P(Y = 4) = \(\frac{1}{3}*\frac{3}{4}\) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
So, the distribution of Y + X: \(\begin{bmatrix}3&4&5&6\\\frac{1}{12}&\frac{1}{3}&\frac{1}{3}&\frac{1}{4}\end{bmatrix}\)
- Y - X
Y - X could be the values {1,2,3,4} For P(Y - X = 1): P(X = 2)P(Y = 3) =
\(\frac{1}{3}*\frac{1}{4}\) = \(\frac{1}{12}\)
For P(Y - X = 2): P(X = 2)P(Y = 4) + P(X = 1)P(Y = 3) = \(\frac{1}{3}*\frac{3}{4}\) + \(\frac{1}{3}*\frac{1}{4}\) = \(\frac{3}{12}\) + \(\frac{1}{12}\) = \(\frac{1}{3}\)
For P(Y - X = 3): P(X = 1)P(Y = 4) + P(X = 0)P(Y = 3) = \(\frac{1}{3}*\frac{3}{4}\) + \(\frac{1}{3}*\frac{1}{4}\) = \(\frac{3}{12}\) + \(\frac{1}{12}\) = \(\frac{1}{3}\)
For P(Y - X = 4): P(X = 0)P(Y = 4) = \(\frac{3}{12}\) = \(\frac{1}{4}\)
So, the distribution of Y - X: \(\begin{bmatrix}1&2&3&4\\\frac{1}{12}&\frac{1}{3}&\frac{1}{3}&\frac{1}{4}\end{bmatrix}\)