Why can’t an elementary particle be forced to have a particular outcome in an entangled pair? I read in a blog Quantum Entanglement: Slower Than Light that one can not force a particle from an EPR pair to have a not statistical outcome for the entanglement parameter. I can not understand why? Is it prohibited by some no go theorem?
A sort of explanation is included in Chad Orzel blog:
He writes there:

If you could measure a particle's state in a way that forced a particular outcome-- you could absolutely send information this way. But you can't do that.

And then:

If Alice does this on her particle, it does not, in fact, affect the state of Bob's particle in any way-- it's still in an indeterminate state that's a mix of 0 and 1.

All this is just a statement and not answer to why the particle can not be forced in a particular state by some process?
 A: The proposed experiment is something like this: Alice and Bob prepare a system in the Bell state:
$$|\psi\rangle = \frac{1}{\sqrt{2}}(|01\rangle  + |10\rangle)$$ 
Alice then goes somewhere and wants to communicate say the value "1" to Bob faster than light. To do this, she acts with the operator $U \otimes \mathbf{1}$ on the state (this is a local operation so she can do this). $U$ is chosen such that if she were now to measure her qubit, she'd measure the value $0$. 
So, first issue: such a unitary doesn't exist. This would need to map $|0\rangle$ and $|1\rangle$ to the same state $|0\rangle$ so is not reversible so can't be unitary*. Okay, this can be worked around by Alice bringing her qubit into contact with a second one (in the state $|0\rangle$ say) and then we can construct a system that maps $|00\rangle \mapsto |00\rangle$ and also $|10\rangle \mapsto |01\rangle$. 
The issue is that none of this has actually affected Bob's qubit in any way. Sure, if Alice now measures her qubit then she will definitely record a $0$. But Bob is still 50% likely to measure either $0$ or $1$. The procedure that Alice used to make her measurements certain involved some fiddly business with a third qubit, and now what is entangled is that third qubit and Bob's.

*The operations of the form $U\otimes \mathbf{1}$ really can't work because they change the state to $|\psi'\rangle = U|0\rangle|1\rangle + U|1\rangle|0\rangle \neq |01\rangle$. Ultimately, there is nothing Alice can do to influence Bob's reduced density matrix other than measuring her qubit. 
