Soft thresholding (Donoho and Johnstone) Donoho and Johnstone (1994) poses the following equality:
$$
E((\eta_t(X) - \mu)^2) = 1 - 2\Pr(|X|\lt t) + E(\min(X^2,t^2))
$$
where $\eta_t(X) = \operatorname{sign}(x)\max(|X|-t,0)$ and $X \sim N(\mu,1)$.
I am having trouble seeing this; any hints?
 A: Think of the function $\eta_t$ as an "additive perturbation" of the identity $\operatorname{Id}:x\to x$ by expressing $\eta_t(X)$ as the difference between $X$ itself and the perturbation $T_t.$  Here is the picture when $t\ge 0:$

This spares us from having to deal with three separate expressions for $\eta_t$ (broken down by whether $x\lt -t,$ $-t\le x\le t,$ or $t \lt x$) which results in up to nine separate expressions for its square. It also suggests that we view $\eta_t(X) - \mu$ as a perturbation of $X-\mu,$ whose expected square is the variance of $X.$  This motivates the first couple of algebraic steps in $(**)$ below.
Before we proceed, let's use the plots to derive some properties of $T_t.$  The function $T_t$ "clamps" $X$ to the range $[-t, t]$ by setting larger values of $X$ to $t$ and smaller values to $-t.$  Notice that $T_t^2$ thereby clamps $X^2$ to the range $[0,t^2];$ that is,
$$T_t(X)^2 = \min(X^2, t^2).$$
For future reference, notice too that $T_t$ is almost everywhere differentiable and its derivative is the indicator of the interval $(-t,t):$
$$T_t^\prime(x) = \frac{d}{dx}T_t(x) = \mathcal{I}_{(-t,t)}(x) = \left\{\matrix{1 & \text{if } -t \lt x \lt t \\ 0 & \text{otherwise.}}\right.\tag{*}$$
After making the substitution
$$\eta_t(X) = X - T_t(X),$$
algebra leads us directly through the next step,
$$\eqalign{
E\left[\left(\eta_t(X) - \mu\right)^2\right] 
&=E\left[\left(X - T_t(X) - \mu\right)^2\right] \\ 
&=E\left[\left(X - \mu - T_t(X)\right)^2\right] \\ 
&=E\left[(X-\mu)^2 - 2(X-\mu)T_t(X) + T_t(X)^2\right] \\
&=E\left[(X-\mu)^2\right] -2 E\left[T_t(X)(X-\mu)\right] + E\left[T_t(X)^2\right]\\
&=\color{red}{\operatorname{Var(X)}} -2 \color{blue}{E\left[T_t(X)(X-\mu)\right]}+ \color{green}{E\left[\min(X^2, t^2)\right]}.\tag{**}
}$$
So far it has been unnecessary to use the assumption that $X$ has a Normal distribution with mean $\mu$ and unit variance. We do so now and apply the Lemma (below) to the function $g(x) = T_t(x)$ and invoke $(*)$ above to re-express the middle term as
$$\color{blue}{E\left[T_t(X)(X-\mu)\right]} = E\left[T_t^\prime(X)\right] = E\left[\mathcal{I}_{(-t,t)}(X)\right] = \Pr(-t \lt X \lt t) = \color{blue}{\Pr(|X| \lt t)}.$$
Along with the assumption $\color{red}{\operatorname{Var}(X) = 1},$ this enables us to express $(**)$ as
$$E\left[\left(\eta_t(X) - \mu\right)^2\right] = \color{red}{1} -2\color{blue}{\Pr(|X| \lt t)} + \color{green}{E\left[\min(X^2, t^2)\right]},$$
QED.
A similar approach will deal with the case of negative $t,$ but the answer will be different.  I surmise the original formula was intended only for non-negative $t.$

Lemma
Let $\phi_\mu$ be the density function of a Normal$(\mu,1)$ variable.  What is special about $\phi_\mu$ is that
$$\frac{\phi_\mu^\prime(x)}{\phi_\mu(x)} = \frac{d}{dx} \log(\phi_\mu(x)) = -(x-\mu).$$
Let $g$ be any integrable and almost everywhere differentiable function.  Inserting the foregoing into the product rule of differentiation yields
$$\frac{d}{dx} \left(g(x)\phi_\mu(x)\right) = g^\prime(x) \phi_\mu(x) + g(x) \phi_\mu^\prime(x) = g^\prime(x) \phi_\mu(x) - g(x)(x-\mu) \phi_\mu(x),$$
so integrating both sides and invoking the Fundamental Theorem of Calculus produces
$$\eqalign{
g(x)\phi_\mu(x) \mid_{-\infty}^\infty &= \int_\mathbb{R}\frac{d}{dx} \left(g(x)\phi_\mu(x)\right)  dx \\
&= \int_\mathbb{R}g^\prime(x) \phi_\mu(x) dx - \int_\mathbb{R}g(x)(x-\mu) \phi_\mu(x) dx 
 \\&= E\left[g^\prime(X)\right] - E\left[g(X)(X-\mu)\right].
}$$
When $g(x)\phi_\mu(x)$ is asymptotically $0$ as $x\to\pm \infty,$ the left side becomes $0,$ providing a very general result for Normally distributed random variables:

$$ E\left[g(X)(X-\mu)\right] = E\left[g^\prime(X)\right] .$$

