Suppose $X\sim N(\mu,1)$, show that $|\mu|$ has no unbiased estimate Suppose $X\sim N(\mu,1)$, show that $|\mu|$ has no unbiased estimator. Hint:use the fact that $|\mu|$ is not differentiable at $\mu=0$. There are my ideas:
Suppose $X_1,\cdots,X_n$ are simple random samples from $N(\mu,1)$, and $g(X_1,\cdots,X_n)$ is an unbiased estimator for $|\mu|$, which means $\mathbb{E}_{\mu}(g(X_1,\cdots,X_n))=|\mu|$, i.e. $$\int g(x_1,\cdots,x_n)e^{-\frac{(x_1-\mu)^2}{2}}\cdots e^{-\frac{(x_n-\mu)^2}{2}}\,\mathrm{d}x_1\cdots\mathrm{d}x_n=|\mu|.$$
I want to prove that $\mathbb{E}_{\mu}(g(X_1,\cdots,X_n))$ is differentiable at $\mu=0$. For this, I want to exchange integral and derivative by Leibniz integral rule on measure theory statement. I am going to find a function $h(x_1,\cdots,x_n)$ that satisfies $\left|\dfrac{\partial G(x_1,\cdots,x_n,\mu)}{\partial \mu}\right|\le h(x_1,\cdots,x_n)$ on $|\mu|\le\epsilon$ ($\epsilon$ is a positive number) and $h(x_1,\cdots,x_n)$ is integrable, where $$G=g(x_1,\cdots,x_n)e^{-\frac{(x_1-\mu)^2}{2}}\cdots e^{-\frac{(x_n-\mu)^2}{2}}.$$ But I am stuck.
 A: $\def\d{\mathrm{d}}\def\R{\mathbb{R}}\def\vec{\boldsymbol}\def\abs#1{\left|#1\right|}\def\paren#1{\left(#1\right)}$For $\vec{x} \in \R^n$ and $μ \in (-1, 1)$,$$
\frac{\partial G}{\partial μ}(\vec{x}, μ) = \exp\paren{ -\frac{1}{2}nμ^2 } · g(\vec{x}) \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 } · \paren{ \sum_{k = 1}^n x_k - nμ } \exp\paren{ μ\sum_{k = 1}^n x_k }.
$$
Since\begin{gather*}
\abs{ \sum_{k = 1}^n x_k - nμ } \leqslant \abs{ \sum_{k = 1}^n x_k } + n \leqslant \exp\paren{ \abs{ \sum_{k = 1}^n x_k } } + n,\\
\exp\paren{ μ\sum_{k = 1}^n x_k } \leqslant \exp\paren{ \abs{ \sum_{k = 1}^n x_k } },
\end{gather*}
then\begin{align*}
\abs{ \frac{\partial G}{\partial μ}(\vec{x}, μ) } &\leqslant |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 } · \abs{ \sum_{k = 1}^n x_k - nμ } \exp\paren{ μ\sum_{k = 1}^n x_k }\\
&\leqslant |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 } · \paren{ \exp\paren{ \abs{ \sum_{k = 1}^n x_k } } + n } \exp\paren{ \abs{ \sum_{k = 1}^n x_k } }\\
&= |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 + 2\abs{ \sum_{k = 1}^n x_k } }\\
&\mathrel{\phantom{=}} + n |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 + \abs{ \sum_{k = 1}^n x_k } }.\tag{1}
\end{align*}
Define\begin{gather*}
F(\vec{x}, s) = |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 + s\abs{ \sum_{k = 1}^n x_k } }\\
D_+ = \left\{ \vec{x} \in \R^n \,\middle|\, \sum_{k = 1}^n x_k \geqslant 0 \right\}, \quad D_- = \left\{ \vec{x} \in \R^n \,\middle|\, \sum_{k = 1}^n x_k < 0 \right\}.
\end{gather*}
Note that for $\vec{x} \in D_+$,\begin{gather*}
F(\vec{x}, s) = |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 + s\sum_{k = 1}^n x_k } = |G(\vec{x}, s)| \exp\paren{\frac{1}{2} ns^2},
\end{gather*}
and for $x \in D_-$,\begin{gather*}
F(\vec{x}, s) = |g(\vec{x})| \exp\paren{ -\frac{1}{2} \sum_{k = 1}^n x_k^2 - s\sum_{k = 1}^n x_k } = |G(\vec{x}, -s)| \exp\paren{\frac{1}{2} ns^2}.
\end{gather*}
Thus\begin{gather*}
\int_{\R^n} F(\vec{x}, s) \,\d \vec{x} = \int_{D_+} F(\vec{x}, s) \,\d \vec{x} + \int_{D_-} F(\vec{x}, s) \,\d \vec{x}\\
\leqslant \exp\paren{\frac{1}{2} ns^2} \paren{ \int_{\R^n} |G(\vec{x}, s)| \,\d \vec{x} + \int_{\R^n} |G(\vec{x}, -s)| \,\d \vec{x} } < +∞,
\end{gather*}
where the last inequality uses the fact that $G(\,·\,, s)$ and $G(\,·\,, -s)$ are integrable.
Now, (1) means that $\abs{ \dfrac{\partial G}{\partial μ}(\vec{x}, μ) } \leqslant F(\vec{x}, 2) + nF(\vec{x}, 1)$. Therefore it follows from the dominated convergence theorem that$$
\left. \frac{\d}{\d μ}\paren{ \int_{\R^n} G(\vec{x}, μ) \,\d \vec{x} } \right|_{μ = 0} = \int_{\R^n} \frac{\partial G}{\partial μ}(\vec{x}, 0) \,\d \vec{x}.
$$
