# Why the correct probability for a square M-QAM is equal to the square of correct probability of ($\sqrt M$)-PAM?

The question is why:

$P_c (\text{square}$ $M-QAM) = (1 - P_{\sqrt M})^2$

where: $P_{\sqrt M}$ is the error probability for $\sqrt M$-PAM signal. Thanks.

The decision in M-QAM is actually two independent sub-decisions:

(i) what is the horizontal coordinate of the constellation point nearest to the received signal point?

and

(ii) what is the vertical coordinate of the constellation point nearest to the received signal point?

Consequently, the decision is correct if and only if both sub-decisions are correct. But the probability of a sub-decision being correct is just $(1-P_{\sqrt M})$, and so the probability that both are correct is $(1-P_{\sqrt M})^2$ by independence of the events. If the I and Q branches of the receiver had mismatched gains, the independence would still hold but the two probabilities being multiplied would be different (corresponding to the different SNRs in the two branches).

There are 2 reasons.

1. Square M-QAM is formed by two $\sqrt{M}$-PAMs. See how they teach multiplication to children, here 4x4 as an example. 1. Noise is isotropic and has equal power and probability of occurrence in all directions.