If we have a sample of $x=2$ from a $Po(6 \cdot \lambda)$ distribution.
How do we calculate $\lambda*$ and $d(\lambda*)$?
I think that $\lambda* = \frac{2}{6} = \frac{1}{3} $ but I am not sure about the other one.
If we have a sample of $x=2$ from a $Po(6 \cdot \lambda)$ distribution.
How do we calculate $\lambda*$ and $d(\lambda*)$?
I think that $\lambda* = \frac{2}{6} = \frac{1}{3} $ but I am not sure about the other one.
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Whatever you may mean by $\lambda *$ and $d(\lambda *),$ which I suppose are defined in your text or notes, the following information may be helpful.
If $X_1$ and $X_2$ are independently $\mathsf{Pois}(6\lambda),$ Then $$S = X_1 + X_2 \sim \mathsf{Pois}(12\lambda).$$ Then $E(S) = 12\lambda$ and $Var(S) = 12\lambda.$
So $S$ is a point estimate of $12 \lambda$ and $S/12$ is a point estimate of $\lambda.$