Distribution of the exponential of a mixture?

Question

Suppose that $X$ is distributed as a finite mixture of normals

$$\sum_{j=1}^k w_j \phi(x;\mu_j,\sigma_j^2).$$

Is $\exp(X)$ distributed as a finite mixture of log-normal distributions?

Answer 1

When operating a change of variable from $X$ to $Y=\exp(X)$, the density gets transformed as follows:$$f_Y(y)=f_X(x\{y\})\times\left|\dfrac{\text{d}x(y)}{\text{d}y}\right|=f_X(\log\{y\})\times\left|\dfrac{\text{d}\log\{y\}}{\text{d}y}\right|=f_X(\log\{y\})=f_X(\log\{y\})\frac{1}{y}$$which translates into$$f_Y(y)=\sum_{j=1}^k w_j \frac{1}{y}\phi(\log\{y\};\mu_j,\sigma_j^2).$$This is a mixture of log-normal densities.

Answer 2

When formulated in the most general way, this result can be made obvious. In full generality the mixture components could be any distributions (not even continuous ones) and the mixture need not even be finite. This approach forces us to reason with the most basic tools available (cumulative distribution functions)--and the conclusion drops right out. I think this gets to the essence of the matter.

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Let $X$ be distributed as a mixture. In terms of cumulative density functions (CDFs) we may write

$$F_X(x) = \sum_i \omega_i F_{i}(x)$$

for weights $(\omega_i)$ which sum to unity and CDFs $F_i$ of independent random variables $X_i$. (More generally, the sum could be an integral.)

Suppose $f$ is a real-valued function defined on a set containing the range of $X$. For example, $f$ might be the exponential function. Applying the definition of a CDF, we obtain the distribution law for any of the $f(X_i)$ as

$$G_i(y) = \Pr(f(X_i)\le y).$$

With these definitions in place, the distribution law for $f(X)$ will be given by

$$G(y) = \Pr(f(X) \le y) = \sum_i \omega_i {\Pr}_i(f(X_i) \le y) = \sum_i \omega_i G_i(y).$$

This is the desired result.

When all the $F_i$ are normal distributions and $f$ is the exponential function, this says the exponential of a mixture of normals is a mixture of lognormals.