Suppose we know nothing about assignment mechanism: what can we say?
Suppose all we know is \(Y^x\in[0,1]\) (w.l.o.g.: with bounds [\(\underline{Y}\), \(\bar{Y}\)] replace \(Y\) by \(\frac{Y-\underline{Y}}{\bar{Y}-\underline{Y}}\))
Theorem: \[\begin{gather*} E[Y^1-Y^0]\in[\left\{E[Y|X=1]P(X=1)-E[Y|X=0](1-P(X=1))\right\}-P(X=1), \\ \left\{E[Y|X=1]P(X=1)-E[Y|X=0](1-P(X=1))\right\}+(1-P(X=1))] \end{gather*}\]
Corollary: Width of possible interval learnable from data is 1 (as opposed to 2 without data) and is \([0,1]\) at largest, \([-1,0]\) at smallest, so worst case interval always contains 0.
Proof: \(E[Y^1-Y^0]=E[Y^1X+Y^1(1-X)]-E[Y^0X+Y^0(1-X)]\) \(=E[Y^1X]-E[Y^0(1-X)]+E[Y^1(1-X)]-E[Y^0X]\)
Have \(E[Y^1X]= E[Y|X=1]P(X=1)\), \(E[Y^0(1-X)]=E[Y|X=0](1-P(X=1))\)
Largest possible effect when \(Y^1=1\) when \(X=0\) and \(Y^0=0\) when \(X=1\), so \(E[Y^1(1-X)]-E[Y^0X]=1-P(X=1)\)
Smallest possible effect when \(Y^1=0\) when \(X=0\) and \(Y^0=1\) when \(X=1\), so \(E[Y^1(1-X)]-E[Y^0X]=P(X=0)\)
Upper and lower intervals in corollary follow when \(E[Y|X=1]=1, E[Y|X=0]=0\) vs \(E[Y|X=1]=0, E[Y|X=0]=1\), respectively
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