Log normal distribution - Where am I wrong? 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.
 A: 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). 
A: 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.
