We start with $P\_0\sim P\_1$ probability measures on $(\Omega, \mathcal{F})$, and we define the new equivalent measure $P\_\alpha=\alpha P\_0+(1-\alpha) P\_1$ with $\alpha\in (0,1)$.

Let $X$ be an integrable random variable mapping to $(\mathbb{R},\mathcal{B})$, and $\mathcal{G}\subset\mathcal{F}$ be a sub-$\sigma$-algebra. Does there exist a $\beta\in(0,1)$ for which
$$\mathbb{E}\_{P\_\alpha}[X\mid\mathcal{G}]=\beta\mathbb{E}\_{P\_0}[X\mid \mathcal{G}]+(1-\beta)\mathbb{E}\_{P\_1}[X\mid\mathcal{G}]?$$

My guess would be $\beta=\alpha$, that would be nice, but I can't prove it. I'm about to present a finance related task that uses such an identity, it's not a homework for me.

I tried using [this](https://math.stackexchange.com/questions/506965/change-of-measure-of-conditional-expectation "this") identity:
$$\mathbb{E}\_P(X\mid\mathcal{G})\mathbb{E}\_Q\left(\frac{dP}{dQ}\mid\mathcal{G}\right)=\mathbb{E}\_Q\left(\frac{dP}{dQ}X\mid\mathcal{G}\right),$$ 
but I only got
$$\beta=1-(1-\alpha)\mathbb{E}\_{P\_\alpha}\left(\frac{dP\_1}{dP\_\alpha}\mid\mathcal{G}\right),$$
which would be fine for my application actually, but it's ugly, I'm afraid I made mistakes. It is equal to my guess iff $\mathbb{E}\_{P\_\alpha}\left(\frac{dP\_1}{dP\_\alpha}\mid\mathcal{G}\right)=1$, and I don't think that it's true in general, $\frac{dP\_1}{dP\_\alpha}$ being independent of $\mathcal{G}$ would imply that.

Any help would be appreciated, thanks!