Log-normal distribution

Question

If $X \sim LogN(\mu,\sigma ^2) $, would the distribution for $aX \sim LogN(\mu+a,\sigma ^2) $ for $ a>0$?

My solution: $log(X) \sim N(\mu,\sigma ^2) \\ log(aX) = log(a) + log(X) \\ log(aX) \sim log(a)+ N(\mu,\sigma ^2)\\ log(aX) \sim N(\mu + log(a),\sigma ^2) \\ aX \sim LogN(\mu+a,\sigma ^2)$

if $a \in \mathbb{R} $, would $X^a \sim LogN(a \mu,\sigma ^2) $? by following the same logic as above?

Answer 1

(I used the $\propto$ Symbol instead of the ~ (tilde) symbol)

$log(X^a) \propto N(\mu, \sigma^2)$

$log(X^a) = log(X)a$

$log(X^a) \propto aN(\mu, \sigma^2)$

Now the scaling of a Gaussian random variable with a scalar $a$ would not only Change the expectation value; it also changes the Standard deviation. It holds: $\mu \rightarrow a \mu$ and also $\sigma^2 \rightarrow a^2 \sigma^2$ because of

$E(a^2 (N(\mu, \sigma^2))^2) - (E(a N(\mu, \sigma^2)))^2 = a^2 (E((N(\mu, \sigma^2))^2) - \mu^2)$ by linearity of $E$.

Hence

$log(X^a) \propto N(a \mu, a^2 \sigma^2)$.

Taking the exponential will shift expectation value and Standard Deviation in a much more complicated way. Therefore it is difficult to compute a log-normal distributed $X^a$.