Let \[X\] be a random variable with distribution \(g(x) = \Sigma_{i=1}^{N}\lambda_{i}f_{i}(x)\), where each \(f_{i}\) is a probability distribution for the RV \(X_{i}\) (with mean \(\mu_{i}\) and variance \(\sigma_{i}^2\)), \(\lambda_{i}\geq0\) and \(\Sigma_{i=1}^{N}\lambda_{i}=1\).
We determine the mean, \(\mathbb{E}[X]=\mu\) and variance, \(\mathbb{V}[X]=\sigma^2\), of \(X\) in terms of those of the \(X_{i}\).
We have
\[ \mu =
\mathbb{E}[X] =
\mathbb{E}[\Sigma_{i=1}^{N}\lambda_{i}X_{i}] =
\Sigma\lambda_{i}\mathbb{E}[X_{i}] =
\Sigma_{i=1}^{N}\lambda_{i}\mu_{i}
\]
by linearity.
Then on expanding the square we have
\[
\sigma^2 = \mathbb{V}[X]=\mathbb{E}[(X-\mu)^{2}]=\mathbb{E}[X^{2}]-\mu^{2}
\],
but that isn't really any help to us, because we haven't determined \( \mathbb{E}[X^2] \).
Alternatively,
\[
\sigma^{2}
= \int dx(x-\mu)^{2}f(x)
= \int dx(x-\mu)^{2}\Sigma_{i}\lambda_{i}f_{i}(x)
= \Sigma_{i}\lambda_{i}\int(x-\mu)^{2}f_{i}(x)
\]
\[
= \Sigma_{i}\lambda_{i}\int dxf_{i}(x)(x-\mu_{i}+\mu_{i}-\mu)^{2}
\]
\[
= \Sigma_{i}\lambda_{i}\int dxf_{i}(x)[(x-\mu_{i})^{2}+2(x-\mu_{i})(\mu_{i}-\mu)+(\mu_{i}-\mu)^{2}]
\]
\[
=\Sigma_{i}\lambda_{i}[\int dxf_{i}(x)(x-\mu_{i})^{2}+2(\mu_{i}-\mu)\int dxf_{i}(x)(x-\mu_{i})+(\mu_{i}-\mu)^{2}\int dxf_{i}(x)]
\]
\[
= \Sigma_{i=1}^{N}\lambda_{i}(\mathbb{E}[(X_{i}-\mu_{i})^{2}]+2(\mu_{i}-\mu)(\mathbb{E}[X_{i}]-\mu_{i})+(\mu_{i}-\mu)^{2})
\]
\[
= \Sigma_{i=1}^{N}\lambda_{i}(\mathbb{V}[X_{i}]+(\mu_{i}-\mu)^{2})
\]
\[
\geq \Sigma_{i} \lambda_{i} \mathbb{V}[X_{i}]
\]
