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Suppose $X=\log(Y)$ can be modeled by a mixture of two normal distributions with proportion $p$ of $X_1$ and proportion $1-p$ of $X_2$, where $X_1\sim\mathcal N(U_1, \sigma^2_1)$ and $X_2\sim\mathcal N(U_2, \sigma^2_2)$.

How do you calculate $E(Y)$; i.e., $E(\exp(X))$ where $X$ is a mixture of two normals?

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    $\begingroup$ Are you asking about a linear combination of normals or a mixture? The notation you've used is a little ambiguous in this regard and a correct answer will depend on which you are interested in. $\endgroup$ – cardinal Aug 16 '12 at 17:58
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    $\begingroup$ @Cardinal Given that the OP has twice used the word "mixture" and did not equate "$p * X_1 + (1-p) * X_2$" with a $\mathcal{N}(p U_1 + (1-p)U_2, p^2 \sigma_1^2 + (1-p)^2 \sigma_2^2)$ distribution, it would be erroneous to interpret this as a linear combination, don't you think? $\endgroup$ – whuber Aug 16 '12 at 18:37
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    $\begingroup$ @whuber: I think that's very highly likely; on the other hand, both answers here have done their calculations assuming (intentionally or inadvertently) the latter rather than the former; and, there was a very recent question where this confusion popped up. So, my response was, shall we say, conservative rather than speculative. :-) $\endgroup$ – cardinal Aug 16 '12 at 18:41
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    $\begingroup$ @Cardinal Thanks. And, to be perfectly clear--because I think you're holding back a little out of politeness--the answer to the mixture interpretation differs substantially from the answer to the linear combination interpretation. $\endgroup$ – whuber Aug 16 '12 at 18:44
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    $\begingroup$ I am sorry not getting back earlier. I mean mixture of normals, not linear combination of two normals. $\endgroup$ – qqq Aug 16 '12 at 21:34
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I'll try to give an answer for the mixture case. Let's formalize the set-up. We consider a random variable $X$ and an indicator random variable $I$, with $P[I=1] = 1-P[I=2] = p$, independent of $X$. Furthermore, for the mixture we have that the law of $X$ given that $I=1$ is the law of $X_1$, which is Gaussian with mean $U_1$ and variance $\sigma^2_1$; and, if $I=2$, the law is that of $X_2$ with law $N(U_2,\sigma^2_2)$.

Then, for $Y=\exp(X)$ we can calculate the expectation as $$ \begin{align} E[Y] &= E[\exp(X)] = p E[\exp(X)|I=1] + (1-p) E[\exp(X)|I=2] \\ &= p E[\exp(X_1)] + (1-p) E[\exp(X_2)] \\ &= p \exp(U_1+ \sigma^2_1/2) + (1-p) \exp(U_2+ \sigma^2_2/2) \end{align} $$ by using the expectation of a log-normal.

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  • $\begingroup$ (+1) There are a couple (inconsequential) typos in the first paragraph. $\endgroup$ – cardinal Aug 16 '12 at 20:51
  • $\begingroup$ Follow up exercise (for the interested reader): Imitate this approach to (easily) generalize to arbitrary finite mixture distributions and arbitrary (Borel measurable) function $g(x)$ in place of $\exp(x)$. $\endgroup$ – cardinal Aug 16 '12 at 20:54
  • $\begingroup$ @cardinal I have corrected some ... don't see any anymore ... but it's 11pm here in Europe ... $\endgroup$ – Ric Aug 16 '12 at 20:58
  • $\begingroup$ I used this initially, but started to doubt whether it is correct. If X ~ mixed normal of X1 and X2 with mixing probability p, does that mean exp(X) ~ mixed lognormal with same mixing probability p? I started to doubt this because this estimator seems to be biased (different from sample mean) for my data. Any suggestions on how to reduce the bias? $\endgroup$ – qqq Aug 16 '12 at 21:42
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    $\begingroup$ @Richard: Sorry guys, my complete story: I need to estimate mean revenue (the population is very heavy tailed). Sample mean is a natural unbiased estimator but has a wide confidence interval. I hope to find an estimator with smaller MSE. I fitted normal mixture to log transformed data. The fit seems good visually. But the point estimator I got (as what Richard suggested before) seems to be different from sample mean. I wonder whether I calculated it correctly and therefore raised the original question. If right, how to reduce the bias? Thanks for your suggestions! $\endgroup$ – qqq Aug 17 '12 at 23:07
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For instance, if you are dealing with a two component mixture, the first moment is calculated as follows: $\mu_{\rm mixture}=\pi_{1} \cdot \mu_{1} + \pi_{2} \cdot \mu_{2}$, where $\pi_{i}\ (i=1,2)$ stands for the components' weights and $\mu_{i}\ (i=1,2)$ for the means.

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  • $\begingroup$ This is true, as far as it goes, but it ignores the non-linear transform (exp) in the problem statement. $\endgroup$ – gung - Reinstate Monica Jul 16 '14 at 18:15

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