Let $X$ be a R.V whose pdf is given by $$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+ (1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$

clearly $X\sim pN(\mu_1,\sigma_1^2)+(1-p)N(\mu_2,\sigma_2^2)=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$

therefore $\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$.

However, if I let $Y=e^X$, then Let $G(y)$ be the cdf of $Y$, then $G(y)=P(Y<y)=P(e^X<y)=p(X<\ln(y))=F_X(\ln(y))$ there fore $g(y)=f(\ln(y))\frac{1}{y}$ so the pdf $g(y)$ of $Y$ is given by

$$g(y)=p\frac{1}{y\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(\ln(y)-\mu_1)^2}{2\sigma_1^2} \right) + (1-p) \frac{1}{y\sqrt{2\pi\sigma_2^2}} \exp\left(-\frac{(\ln(y)-\mu_2)^2}{2\sigma_2^2}\right)$$

i.e $Y=p\times \mathrm{logNormal}(\mu_1,\sigma_1^2)+(1-p)\times \mathrm{logNormal} (\mu_2,\sigma_1^2)$

where $\mathrm{logNormal}(\mu_1,\sigma_1^2)$ means a log-normal distribution.

hence $\kappa=E(Y)-1=pE(\mathrm{logNormal}(\mu_1,\sigma_1^2))+(1-p) E(\mathrm{logNormal} (\mu_2,\sigma_2^2))= pe^{\mu_1+\frac{\sigma_1^2}{2}}+(1-p)e^{\mu_2+\frac{\sigma_2^2}{2}}-1$.

My question is : why I have different values for $\kappa$? Could some one explain for me where I am wrong ? thank you for your time.

  • $\begingroup$ The first part of your post is incorrect. The distribution of a mixture is not the distribution of a weighted sum. $\endgroup$ – user76844 Aug 28 '14 at 16:15
  • $\begingroup$ Your second line is incorrect. $X$ has a mixture normal distribution and not a normal distribution as the last equality on the second line asserts. Your calculation of the pdf of $Y$ as a mixture of log-normal distributions is correct; the statement made on line 3 is not. $\endgroup$ – Dilip Sarwate Aug 28 '14 at 16:15
  • $\begingroup$ Eupraxix1981: why is that ? so what is the distribution of $X$ ? I found it here en.wikipedia.org/wiki/… $\endgroup$ – Dave Nguyen Aug 28 '14 at 16:20
  • $\begingroup$ Dilip : isn't if $Y\sim logNormal(\mu,\sigma^2)$ then $E(Y)=e^{\mu+0.5\sigma^2}$? $\endgroup$ – Dave Nguyen Aug 28 '14 at 16:22

With $p$ denoting a constant in the interval $(0,1)$, suppose $X$ has density $$f(x)=p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+ (1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)$$ which is a mixture density. Then, $$\begin{align} F_X(\alpha) &= P\{X \leq \alpha\}\\ &=\int_{-\infty}^{\alpha} f(x)\,\mathrm dx\\ &= \int_{-\infty}^{\alpha}p\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)+ (1-p)\frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)\,\mathrm dx\\ &= p\int_{-\infty}^{\alpha}\frac{1}{\sqrt{2\pi\sigma_1^2}}\exp\left(-\frac{(x-\mu_1)^2}{2\sigma_1^2}\right)\,\mathrm dx+(1-p)\int_{-\infty}^{\alpha} \frac{1}{\sqrt{2\pi\sigma_2^2}}\exp\left(-\frac{(x-\mu_2)^2}{2\sigma_2^2}\right)\,\mathrm dx\\ &= p\Phi\left(\frac{\alpha-\mu_1}{\sigma_1}\right) +(1-p)\Phi\left(\frac{\alpha-\mu_1}{\sigma_1}\right)\\ &\neq \Phi\left(\frac{\alpha-(p\mu_1+(1-p)\mu_2)}{\sqrt{p^2\sigma_1^2+(1-p)^2\sigma_2^2}}\right) \end{align}$$

In other words, it is not true (as you state on Line 3 of your question, and continue to assert dogmatically) that $X$ has a normal distribution with mean $p\mu_1+(1-p)\mu_2)$ and variance $p^2\sigma_1^2+(1-p)^2\sigma_2^2$. That is, $$X\sim N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)\quad \mathbf{is~a~false~statement}$$

As a result, your calculation $\kappa=E(e^X)-1=e^{p\mu_1+(1-p)\mu_2+\frac{1}{2}(p^2\sigma_1^2+(1-p)^2\sigma_2^2)}-1$ is not correct.

  • $\begingroup$ Thanks Dilip :)) I got it. $\endgroup$ – Dave Nguyen Aug 29 '14 at 15:10

Let $\mu := p\mu_1+(1-p)\mu_2$ and $\sigma^2 := p^2\sigma_1^2 + (1-p)^2\sigma_2^2$. Then, one can show that, the pdf of $Y$ is: $$g(y)=\frac{1}{y\sqrt{2\pi\sigma^2}}\exp\left(-\frac{(\ln(y)-\mu)^2}{2\sigma^2}\right),$$ not the one that you display (your $Y$ presentation as linear combination of log-normal variables is wrong: if $X=pX_1 +(1-p)X_2$, then $Y={\rm e}^X = {\rm e}^{pX_1}{\rm e}^{(1-p)X_2}$, not $p{\rm e}^{X_1}+(1-p){\rm e}^{X_2}$; also, $f$, pdf of $X=pX_1+(1-p)X_2$, is NOT $pf_1 + (1-p)f_2$, where $f_1$ and $f_2$ are pdf's of $X_1$ and $X_2$; lastly, see Dilip Sarwarte comment on what is really your starting hypothesis on $X$ being normal variable, as say a linear combination of independent normal variables, or on its pdf being a mixture of normal pdf's).

  • $\begingroup$ But what you wrote is exactly $X=N(p\mu_1+(1-p)\mu_2,p^2\sigma_1^2+(1-p)^2\sigma_2^2)$? I am confused now :(( Here is what I have Let $G(y)$ be the cdf of $Y$, then $G(y)=P(Y<y)=P(e^X<y)=p(x<ln(y))=F_X(ln(y))$ there fore $g(y)=f(ln(y))\frac{1}{y}$ $\endgroup$ – Dave Nguyen Aug 28 '14 at 16:26
  • $\begingroup$ Yes. $Y={\rm e}^X$ and $X$ is normal with params $\mu$ and $\sigma^2$. Hence $Y$ is log-normal. Hence $g$ form. $\endgroup$ – ir7 Aug 28 '14 at 16:29
  • $\begingroup$ Here is what I have Let $G(y)$ be the cdf of $Y$, then $G(y)=P(Y<y)=P(e^X<y)=p(X<ln(y))=F_X(ln(y))$ there fore $g(y)=f(ln(y))\frac{1}{y}$ $\endgroup$ – Dave Nguyen Aug 28 '14 at 16:31
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    $\begingroup$ $f$, pdf of $X=pX_1+(1-p)X_2$, is NOT $pf_1 + (1-p)f_2$, where $f_1$ and $f_2$ are pdf's of $X_1$ and $X_2$. $\endgroup$ – ir7 Aug 28 '14 at 16:48
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    $\begingroup$ @Nguyen That's ok. In my answer, I mentioned Dilip Sawarte's comment. Basically, if you stop claiming that $X$ is normal, then $f=pf_1+(1-p)f_2$, as given directly by the problem. Hence, your $g$ is correct. If you claim that $X$ is normal (with $\mu$ and $\sigma$), then my $g$ is correct. $\endgroup$ – ir7 Aug 28 '14 at 19:15

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