Yes, there will be payback to the world, but not as much as if you spend the money on efficient charity (I would think).
Why do you think that? Here’s a simple model. Suppose that the value of what you do and create (excluding offspring) in your life is A. You are paid some amount B for this. (If you have an honest job, B is less than A.) Of that, you spend C on your own household excluding children, leaving B-C for either charitable giving or raising offspring. Let D be the cost of raising a child. The number you can afford to raise is N = (B-C)/D.
If you do not have a child, the value of your life to others is A-C.
If you have K children, the value of all your lives (assuming the same figures apply to your children) is (K+1)(A-C-KD).
The former minus the latter is (K+1)KD-K(A-C). This favours not reproducing only if K > (A-C)/D − 1. But suppose you have one less child than the maximum you can afford. Then K = N-1 = (B-C)/D − 1. If you have an honest job, this must be less than (A-C)/D − 1. The optimum value of K turns out to be ((A-C)/D − 1)/2 (or some value a little less than N, if that is smaller).
So the conclusion of this calculation is “be fruitful and multiply”.
Why do you think that? Here’s a simple model. Suppose that the value of what you do and create (excluding offspring) in your life is A. You are paid some amount B for this. (If you have an honest job, B is less than A.) Of that, you spend C on your own household excluding children, leaving B-C for either charitable giving or raising offspring. Let D be the cost of raising a child. The number you can afford to raise is N = (B-C)/D.
If you do not have a child, the value of your life to others is A-C.
If you have K children, the value of all your lives (assuming the same figures apply to your children) is (K+1)(A-C-KD).
The former minus the latter is (K+1)KD-K(A-C). This favours not reproducing only if K > (A-C)/D − 1. But suppose you have one less child than the maximum you can afford. Then K = N-1 = (B-C)/D − 1. If you have an honest job, this must be less than (A-C)/D − 1. The optimum value of K turns out to be ((A-C)/D − 1)/2 (or some value a little less than N, if that is smaller).
So the conclusion of this calculation is “be fruitful and multiply”.