Given the following deterministic equivalent definition of two-stage stochastic linear problems:
$K = min_{x} \left\{c^Tx + \sum_{k=1}^{M} p_kq_k^Ty_k |\ Ax = b; \ T_kx + W_ky_k = h_k; y_k\geq 0, x\geq 0 \,\,\forall\ k=1,2,\ldots M \right\}$
where $(T_k, W_k, h_k, y_k)$ occurs with probability $p_k > 0$ $\ \forall\ k=1,2,\ldots, M$
Note that the term $\sum_{k=1}^{M} p_kq_k^Ty_k$ with the condition $T_kx + W_ky_k = h_k; y_k\geq 0$ is actually equivalent to the second-stage problem in SLP:
$E(Q(x,\zeta))$, where $Q(x,\zeta) = min_{y} \left\{q^Ty | Tx + Wy = h, y\geq 0\right\}$, and $\zeta = (q,T,w,h)$.
Now, I want to prove the following claim: For $M\geq 2$, $$K\neq p_1 min_{x} \left\{c^Tx + q_1^Ty_1 |\ Ax = b; \ T_1x + W_1y_1 = h_1; y_1\geq 0, x\geq 0\right\} + \ldots + p_M min_{x} \left\{c^Tx + q_M^Ty_M |\ Ax = b; \ T_Mx + W_My_M = h_M; y_M\geq 0, x\geq 0\right\}$$
1st approach:
First, since $\sum_{k=1}^{M}p_k = 1$, I planned to establish $min_{x} \left\{c^Tx + \sum_{k=1}^{M} p_kq_k^Ty_k |\ Ax = b; \ T_kx + W_ky_k = h_k; y_k\geq 0, x\geq 0 \,\,\forall\ k=1,2,\ldots M \right\}> \sum_{k=1}^{M} min_{x} \left\{p_kc^Tx + p_kq_k^Ty_k |\ Ax = b; \ T_kx + W_ky_k = h_k; y_k\geq 0, x\geq 0\right\}$
However, such inequality seems not to always hold, as we know nothing about the signs of the coefficients of the objective function, besides the fact that the constraints form a polyhedron as a feasible region. The reverse sign is also not correct for every cases, I think. So I abandon this approach.
2nd approach:
Define optimal solutions $x^{*}$ for K, and $x_{1}^{*},x_{2}^{*}, \ldots, x_{M}^{*}$ for $min_{x} \left\{p_kc^Tx + p_kq_k^Ty_k |\ Ax = b; \ T_kx + W_ky_k = h_k; y_k\geq 0, x\geq 0\right\}$ $\forall \ k=1,2,\ldots, K$.
By the sake of contradiction, assume $K = \sum_{k=1}^{M} min_{x} \left\{p_kc^Tx + p_kq_k^Ty_k |\ Ax = b; \ T_kx + W_ky_k = h_k; y_k\geq 0, x\geq 0\right\}$. This is equivalent to:
$$c^Tx^{*} = \sum_{k=1}^{M} p_kc^Tx_k^{*}$$.
However, note that $Ax^{*} = Ax_{1}^{*} = Ax_{2}^{*} =\ldots = Ax_{M}^{*} (\ =\ b)$, and $T_kx^{*} = T_kx_{1}^{*} =\ldots = T_kx_{M}^{*}$. So, does the first set of equality imply $x^{*} = x_{1}^{*} = x_{2}^{*} = \ldots = x_{M}^{*}$, if $A$ is an invertible matrix?? But if that's the case, shouldn't this means $\sum_{k=1}^{M} p_kc^Tx_k^{*} = c^Tx^{*}\sum_{k=1}^{M} p_k = c^Tx^{*}$ (since $\sum_{k=1}^{M}p_k = 1$)? But this implies the claim above is wrong...
My question: Could someone please help review my attempt above, and let me know how you could arrive at the desired contradiction, or if the claim is actually false? Any other suggestions for other approaches would be really appreciated.