Question

Suppose that a random variableXhas the normal distributionwith meanμand precision\(\tau \). Show that the random variable\({\bf{Y = aX + b}}\;\left( {{\bf{a}} \ne {\bf{0}}} \right)\)has the normal distribution with meanaμ+band precision\(\frac{\tau }{{{{\bf{a}}^{\bf{2}}}}}\).

Short answer

Proved. The random variable \(Y = aX + b\;\left( {a \ne 0} \right)\) has a normal distribution with mean \(a\mu + b\) and precision \(\frac{\tau }{{{a^2}}}\).

Step-by-step solution

Given information

X is a random variable from the normal distribution with mean\(\mu \)and precision\(\tau \). Consider a new variable Y. \(Y = aX + b\).

describe the mean and variance of the new variable

As there can be concluded that if X is a random variable that follows a normal distribution with mean\(\mu \)and variance\({\sigma ^2}\)then the random variable\(Y = aX + b\;\left( {a \ne 0} \right)\)has the normal distribution with mean\(a\mu + b\)and variance\({a^2}{\sigma ^2}\).

As,

\(\begin{align}E\left( Y \right) &= aE\left( X \right) + E\left( b \right)\\ &= a\mu + b\end{align}\)

And

\(\begin{align}Var\left( Y \right) &= {a^2}Var\left( X \right) + 0\\ &= {a^2}{\sigma ^2}\end{align}\)

Calculate the precision

By the definition of precision, a precision of a normal distribution can be described as the reciprocal of the variance of the variable.

Therefore, for the random variable\(Y = aX + b\;\left( {a \ne 0} \right)\), the variance is\({a^2}{\sigma ^2}\). So, the reciprocal can be expressed as\(\frac{1}{{{a^2}{\sigma ^2}}}\).

Now for the X variable, the precision is\(\tau = \frac{1}{{{\sigma ^2}}}\)

So, the precision of Y is, \(\frac{\tau }{{{a^2}}}\)and the mean is \(a\mu + b\).