Question

Let the random variable \(X\) have a distribution that is \(N\left(\mu, \sigma^{2}\right)\). (a) Does the random variable \(Y=X^{2}\) also have a normal distribution? (b) Would the random variable \(Y=a X+b, a\) and \(b\) nonzero constants have a normal distribution? Hint: \(\operatorname{In}\) each case, first determine \(P(Y \leq y)\).

Short answer

(a) No, \(Y = X^{2}\) will not follow a normal distribution. (b) Yes, \(Y= aX + b\) will have a normal distribution.

Step-by-step solution

Analyze the distribution of squared random variable

For the random variable \(Y\) where \(Y = X^2\), the square of a normal random variable does not follow a normal distribution unless the original variable is always positive or always negative. This is because when a random variable is squared, the negative part of its distribution will become positive, making the distribution non-symmetric, and thus, not normal.

Determine cumulative distribution function of \(Y = X^2\)

The cumulative distribution function \(P(Y \leq y)\) would be \(P(X^2 \leq y)\). This is a chi-square distribution if \(X\) has a standard normal distribution. But in this case we were given the distribution of \(X\) as \(N(\mu, \sigma^{2})\), so \(X^2\) does not follow a chi-square distribution. Thus, \(X^2\) does not have a normal distribution.

Analyze the distribution of linear transformation of the random variable

For the random variable \(Y\) where \(Y = aX + b\) where \(a\) and \(b\) are non-zero constants, the linear transformation of a normal random variable will also follow a normal distribution. This is because linear transformations (addition, subtraction, multiplication, and division by constants) of a normally distributed variable will always be normal due to the closure property of normal distribution.

Determine the cumulative distribution function of \(Y = aX + b\)

The cumulative distribution function \(P(Y \leq y)\) would be \(P(aX + b \leq y)\), which can be re-arranged to \(P(X \leq (y-b)/a)\). This is same form as the cumulative distribution function of \(X\). Thus, \(Y = aX + b\) has a normal distribution.