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”.
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”.