Consider the random variables $Y, Z_1, Z\equiv(Z_1,\dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that $$\mathbb E\left(\frac{Y}{Z_1}\right)=\mathbb E\left(\frac{\mathbb E(Y\mid Z)}{Z_1} \right)$$?
Does the relation hold if we remove i.i.d.?
Consider the random variables $Y, Z_1, Z\equiv(Z_1,\dots,Z_n)$ with $Z_1,...,Z_n$ i.i.d.
Is it true that $$\mathbb E\left(\frac{Y}{Z_1}\right)=\mathbb E\left(\frac{\mathbb E(Y\mid Z)}{Z_1} \right)$$?
Does the relation hold if we remove i.i.d.?
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Yes. Since $\frac 1 {Z_1}$ is measurable with respect to $\sigma (Z_1,Z_2,..,Z_n)$ we can write the right side as $$EE\left(\frac Y {Z_1}\mid Z\right)=E\left(\frac Y {Z_1}\right)$$