Let $T_{i}$ be a random variable describing the survival time for the $i$th sample.
This random variable may be modelled using a conventional linear model
$T_{i} = e^{X'_{i}\beta + \epsilon_{i}}$ (1)
where $\epsilon_{i}$ is the error term for the ith sample.
It's logarithm form is
$\log(T_{i}) = X'_{i}\beta + \epsilon_{i}$
In a text I am looking at, it says,
'if the $\epsilon_{i}$ are normally distributed, then, one obtains a log - normal model for the $T_{i}$'.
From Wikipedia, https://en.wikipedia.org/wiki/Log-normal_distribution
it is mentioned that for a log - normal random variable, X, this random variable may be described as
$X = e^{Z\sigma + \mu}$(2)
I fail to see the connection between (1) and (2). Any help is appreciated.
where $\sigma$ is the standard deviation, $\mu$ is the mean an $Z$ is the standard normal variable.
From $X = e^{Z\sigma + \mu}$ you have that $\ln(X) = \mu + \sigma Z$. Now, $$\text{E}(\ln(X)) = \text{E}(\mu + \sigma Z) = \mu$$ because $\mu$ and $\sigma$ are constant and $\text{E}(Z)=0$.
Also, we have that $$\text{Var}(\ln(X)) = \text{Var}(\mu + \sigma Z) = \sigma^{2} \text{Var}(Z)= \sigma^{2}$$ because $\text{Var}(Z)=1$ and because of the properties of the variance.
From the fact that $Z$ is a standard normal, and $X$ is a linear function of $Z$, we have that $\ln(X)$ is a normal random variable.
So, from $X = e^{Z\sigma + \mu}$ with $Z$ a standard normal, you have that $\ln(X)$ is a normal random variable with mean $\mu$ and variance $\sigma$. You can also do this in reverse, obtaining the former statement from the latter.