Answer:
You have defined two new random variables, \(u\) and \(v\), based on the independent normal random variables \(x_1\), \(x_2\), and \(x_3\).
1. For \(u = x_1x_2x_3\):
- To find the probability distribution of \(u\), you can use the properties of the product of independent random variables. Since \(x_1\), \(x_2\), and \(x_3\) are independent and normally distributed, the product \(u = x_1x_2x_3\) would follow a normal distribution as well.
- The mean of \(u\) would be the product of the means of \(x_1\), \(x_2\), and \(x_3\), which is \(1 * 1 * 1 = 1\).
- The variance of \(u\) would be the product of the variances of \(x_1\), \(x_2\), and \(x_3\), which is \(1 * 1 * 1 = 1\).
So, \(u\) would be normally distributed with a mean of 1 and a variance of 1.
2. For \(v = x_1(2x_2)(3x_3) = 6x_1x_2x_3\):
- Similar to the previous case, \(v\) is also a product of independent random variables \(x_1\), \(x_2\), and \(x_3\), so it follows a normal distribution.
- The mean of \(v\) would be \(6 * 1 = 6\) (since each \(x_i\) has a mean of 1).
- The variance of \(v\) would be the product of the variances of \(x_1\), \(x_2\), and \(x_3\), which is \(1 * 1 * 1 = 1\).
So, \(v\) would be normally distributed with a mean of 6 and a variance of 1.
In summary, both \(u\) and \(v\) are normally distributed random variables. \(u\) has a mean of 1 and a variance of 1, while \(v\) has a mean of 6 and a variance of 1.