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.