0
$\begingroup$

Suppose we are given $X \sim \mathcal{N}(\mu,\Sigma)$. Then, we define the random variable $Y$ as follows:

$Y_i = 1 + X_i $ if $X_i \ge 0$

$Y_i = \exp(X_i)$ if $X_i \lt 0$.

How do I go about calculating the probability density of Y? And $E[Y_i]$ for all $i$? (not a homework problem - it is taken from a paper where the Y model the prior distribution of surface emissions of methane...)

Is the derivation in @martini's reply correct if the $X$ are correlated ($\Sigma$ is not diagonal)?

$\endgroup$
1
$\begingroup$

For the density: Note that $X_i \sim N(\mu_i, \Sigma_{ii})$. Hence, for $t \in \mathbf R^+$, $t \le 1$, we have \begin{align*} \def\P{\mathbf P}\P(Y_i \le t) &= \P(e^{X_i} \le t)\\ &= \P(X_i \le \log t)\\ &= \frac 1{\sqrt{2\pi\Sigma_{ii}}} \int_{-\infty}^{\log t} \exp\bigl(-(x-\mu_i)^2/2\Sigma_{ii}\bigr)\, dx \end{align*} For $t \ge 1$, we have \begin{align*} \P(Y_i \le t) &= \P(Y_i \le 1) + \P(1 \le Y_i \le t)\\ &= \P(Y_i \le 1) + \P(0 \le X_i \le t-1)\\ &= \P(X_i \le 0) + \P(0 \le X_i \le t-1)\\ &= \P(X_i \le t-1)\\ &= \frac 1{\sqrt{2\pi\Sigma_{ii}}} \int_{-\infty}^{t-1} \exp\bigl(-(x-\mu_i)^2/2\Sigma_{ii}\bigr)\, dx \end{align*} Taking derivatives, we say that $Y_i$'s density is given by $$ f_i(t) = \begin{cases} 0 & t \le 0\\ \frac 1t \cdot (2\pi\Sigma_{ii})^{-1/2} \exp\bigl(-(\log t - \mu_i)^2/2\Sigma_{ii}\bigr) & 0< t \le 1\\ (2\pi \Sigma_{ii})^{-1/2}\exp\bigl(-(t-1 - \mu_i)^2/2\Sigma_{ii}\bigr) & t > 1 \end{cases} $$ For the expectation, compute $\int_{\mathbf R} tf_i(t)\, dt$.

$\endgroup$
  • $\begingroup$ great, thanks. one comment: you seem to have left out the "-" sign inside the $exp$ functions. $\endgroup$ – qazxswedc May 4 '15 at 10:19
  • $\begingroup$ i'm not sure if this derivation works if $\Sigma$ is not diagonal. If the $X_i$ are correlated, should there not be some correlation between the $Y_i$ as well? $\endgroup$ – qazxswedc Oct 22 '15 at 8:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.