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Henry
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From your first two lines, your prior implies $E[Y] = N\frac{a}{b}$

So with $E(\lambda\mid Y)=\frac{Y+a}{N+b}$, you have $E\left[E(\lambda\mid Y)\right] =E\left[\frac{Y+a}{N+b}\right]\frac{=E\left[Y\right] +a}{N+b}= \frac{N\frac{a}{b}+a}{N+b} = \frac{a}{b}$$E\left[E(\lambda\mid Y)\right] =E\left[\frac{Y+a}{N+b}\right]=\frac{E\left[Y\right] +a}{N+b}= \frac{N\frac{a}{b}+a}{N+b} = \frac{a}{b}$

which is what you might have thought from your first line and the law of total expectation

From your first two lines, your prior implies $E[Y] = N\frac{a}{b}$

So with $E(\lambda\mid Y)=\frac{Y+a}{N+b}$, you have $E\left[E(\lambda\mid Y)\right] =E\left[\frac{Y+a}{N+b}\right]\frac{=E\left[Y\right] +a}{N+b}= \frac{N\frac{a}{b}+a}{N+b} = \frac{a}{b}$

which is what you might have thought from your first line and the law of total expectation

From your first two lines, your prior implies $E[Y] = N\frac{a}{b}$

So with $E(\lambda\mid Y)=\frac{Y+a}{N+b}$, you have $E\left[E(\lambda\mid Y)\right] =E\left[\frac{Y+a}{N+b}\right]=\frac{E\left[Y\right] +a}{N+b}= \frac{N\frac{a}{b}+a}{N+b} = \frac{a}{b}$

which is what you might have thought from your first line and the law of total expectation

Source Link
Henry
  • 42.2k
  • 1
  • 81
  • 146

From your first two lines, your prior implies $E[Y] = N\frac{a}{b}$

So with $E(\lambda\mid Y)=\frac{Y+a}{N+b}$, you have $E\left[E(\lambda\mid Y)\right] =E\left[\frac{Y+a}{N+b}\right]\frac{=E\left[Y\right] +a}{N+b}= \frac{N\frac{a}{b}+a}{N+b} = \frac{a}{b}$

which is what you might have thought from your first line and the law of total expectation