Is the mean of the truncated normal distribution monotone in $\mu$?

I am wondering whether the mean of the truncated normal distribution is always increasing in $\mu$. The untruncated distribution of $x$ is $\mathcal{N}(\mu,\sigma^2)$. The mean of the truncated distribution is given by $$E[x|a\le x\le b]=\mu+\frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)}\sigma,$$ where $\alpha=(a-\mu)/\sigma, \beta=(b-\mu)/\sigma$. Taking the derivative and simplifying somewhat, we get $$\frac{\partial E[x|a\le x\le b]}{\partial \mu}=1+\frac{[-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2}{[\Phi(\beta)-\Phi(\alpha)]^2}.$$ Thus, for example, if $\phi(\beta)>>\phi(\alpha)$ the last term in the numerator is negative, but then $[-\phi'(\alpha)+\phi'(\beta)]$ should be positive, since $\alpha$ must be in the left tail while $\beta$ is close to the mean.

I am not sure if there are $a,b,\mu,\sigma$ such that this derivative can become negative. In some numerical calculations I did not find such a profile. Can you find an example, or can you instead prove that it is always nonnegative?

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To "address all the concerns" we prove two results: first that the derivative of the truncated mean w.r.t. $\mu$ has an upper bound of unity. Second that this derivative is equal to the ratio of the truncated variance over the untruncated variance (and hence is always positive). $$\text{Result A:} \frac{\partial E[x|a\le x\le b]}{\partial \mu} \le 1$$

We will use results related to log-concave functions (i.e. functions whose logarithm is a concave function). Please see link and references therein. Obviously, $\beta\,\gt\alpha$. Define then the non-negative bi-variate function $F(\beta,\alpha)\equiv {\Phi(\beta)-\Phi(\alpha)}$. It can be equivalently written as $$F(\beta,\alpha)\,=\,\int_{\alpha}^{\beta} \phi(\tau)d\tau=\,\int_{-\infty}^{\infty} \phi(\tau)I(\alpha\le\tau\le \beta)d\tau$$ where $I()$ is the indicator function. The indicator function is log-concave. The standard normal pdf $\phi$ is log-concave. The product of two log-concave functions is log-concave. If we integrate a log-concave function over one of its arguments, the resulting function is log-concave w.r.t. to the remaining variables. So $F(\beta,\alpha)$ is log-concave in $(\beta,\alpha)$ (this is a reproduction of a proof found in Pratt, J. W. (1981). Concavity of the log likelihood. Journal of the American Statistical Association, 76(373), 103-106.) Consider now the uni-variate function $$\ln H(\mu)=\ln F(\beta(\mu),\alpha(\mu))$$ Note that both $\beta$ and $\alpha$ are linear functions of $\mu$. Then $\ln H(\mu)$ is a concave function in $\mu$ (see for example p. 86 eq. [3.15], Boyd & Vandenberghe (2004). Convex optimization, noting that the 2nd derivatives of $\beta$ and of $\alpha$ w.r.t. $\mu$ are zero). Now

$$\frac{\partial \ln H(\mu)}{\partial \mu}\,=\,\frac{\phi(\beta)-\phi(\alpha)}{\Phi(\beta)-\Phi(\alpha)} \left(\frac{-1} {\sigma}\right)= \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)} \left(\frac{1} {\sigma}\right)$$ Multiply throughout by $\sigma^2:$ $$\sigma^2\frac{\partial \ln H(\mu)}{\partial \mu}\,= \frac{\phi(\alpha)-\phi(\beta)}{\Phi(\beta)-\Phi(\alpha)} \sigma$$

Then we can write

$$E[x|a\le x\le b]=\mu+\sigma^2\frac{\partial \ln H(\mu)}{\partial \mu}\tag{1}$$

and therefore

$$\frac{\partial E[x|a\le x\le b]}{\partial \mu}=1+\sigma^2\frac{\partial^2 \ln H(\mu)}{\partial \mu^2}\tag{2}$$

Since $\ln H(\mu)$ is concave, its second derivative is non-positive. So we have established: $$\frac{\partial E[x|a\le x\le b]}{\partial \mu}\le 1 \qquad \forall \;(\mu,\sigma,a,b, a \lt b) \tag{3}$$

Intuitively, the mean of the truncated distribution never changes as much as the underlying location parameter (don't expect the equality to hold).

$$\text{Result B:} \frac{\partial E[x|a\le x\le b]}{\partial \mu}=\frac{\operatorname{Var}_{tr}(x)}{\sigma^2}\gt 0$$

For compactness we will use the following shorthands: ${\Phi(\beta)-\Phi(\alpha)}\equiv Z$, which is a function of $\mu$, $E[x|a\le x\le b]\equiv E_{tr}(x)$, $\frac{\partial \ln H(\mu)}{\partial \mu}\equiv h'$. Expressing the truncated mean in integral form we have $$E_{tr}(x)=\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx$$

$$\Rightarrow \frac{\partial E_{tr}(x)}{\partial \mu} = \int_a^bx\frac{1}{Z\sigma}\phi'\left(\frac{x-\mu}{\sigma}\right)\left(\frac{-1}{\sigma}\right)dx\;+\;\left[\phi(\beta)-\phi(\alpha)\right]\frac{1}{Z\sigma}\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx$$

Now $$\phi'\left(\frac{x-\mu}{\sigma}\right)=(-1)\left(\frac{x-\mu}{\sigma}\right)\phi\left(\frac{x-\mu}{\sigma}\right)$$ Also, the last integral equals $E_{tr}(x)$, while $\left[\phi(\beta)-\phi(\alpha)\right]\frac{1}{Z\sigma}=-h'$. So we have

$$\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\int_a^bx\frac{1}{Z\sigma}\left(x-\mu\right)\phi\left(\frac{x-\mu}{\sigma}\right)dx\;-\;h'E_{tr}(x) \tag{4}$$

Denote the remaining integral $I$ and break it in two: $$\left(\frac{1}{\sigma^2}\right)I=\left(\frac{1}{\sigma^2}\right)\int_a^bx^2\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx\;-\;\left(\frac{\mu}{\sigma^2}\right)\int_a^bx\frac{1}{Z\sigma}\phi\left(\frac{x-\mu}{\sigma}\right)dx$$ The first integral is the second raw moment of the truncated distribution, while the second is $E_{tr}(x)$. So $$\left(\frac{1}{\sigma^2}\right)I=\left(\frac{1}{\sigma^2}\right)E_{tr}(x^2)\;-\;\left(\frac{\mu}{\sigma^2}\right)E_{tr}(x)$$ Inserting into eq.(4) we have

$$\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)E_{tr}(x^2)\;-\;\left(\frac{\mu}{\sigma^2}\right)E_{tr}(x)\;-\;h'E_{tr}(x)$$

$$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\mu E_{tr}(x)\;-\;h'\sigma^2E_{tr}(x)\right]$$

$$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\left(\mu +h'\sigma^2\right)E_{tr}(x)\right]$$

From eq.$(1)$ we have $\mu +h'\sigma^2=E_{tr}(x)$. Substituting we obtain $$\Rightarrow\frac{\partial E_{tr}(x)}{\partial \mu} = \left(\frac{1}{\sigma^2}\right)\left[E_{tr}(x^2)\;-\;\left(E_{tr}(x)\right)^2\right]=\frac{\operatorname{Var}_{tr}(x)}{\sigma^2}\gt 0$$

which is what we wanted to prove.

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Very nice. Very elegantly shown. I didn't get why $\text{ln}~H(\mu)$ is always concave in $\mu$ though (since $F(\beta,\alpha)=\Phi(\beta)-\Phi(\alpha)$ is not linear in $\mu$). – Nameless Jul 30 '13 at 22:33
Thanks for the good words. The arguments alpha and beta of F are linear functions of mu. Then concavity (or convexity) is "inherited" from (alpha, beta) to mu. This is a standard result, you can look it up everywhere, this is why I only gave a reference and did not prove it in detail. – Alecos Papadopoulos Jul 31 '13 at 1:16
Ok, this answers the question and everything seems to be ok. Unless someone else finds a mistake, you deserve the bounty. Nice work. – Nameless Jul 31 '13 at 18:34
Mistake, there isn't, I am a.s. certain (i.e. except maybe for some events with probability zero). But perhaps somebody else will prove the same result using an unexpected method, in which case (s)he should get the bounty. – Alecos Papadopoulos Jul 31 '13 at 19:57

This is not a full answer. I thought I'd try and see how far one could get for an arbitrary probability distribution function $f$.

Truncated to $[a,b]$, the mean is $$m=\frac{\int_a^bxf(x)\,\mathrm dx}{\int_a^bf(x)\,\mathrm dx}=:\frac pq,$$ with \begin{align} p&=\int_a^bxf(x)\,\mathrm dx,\\ q&=\int_a^bf(x)\,\mathrm dx. \end{align} We want to translate $f$ to the right by some $\mu$. Shifting the origin to $\mu$, this is equivalent to translating the bounds $[a,b]$ to the left, and we just have to add $\mu$ to the new mean to account for the change in origin. So $$m(\mu) = \mu + \frac{\int_{a-\mu}^{b-\mu}xf(x)\,\mathrm dx}{\int_{a-\mu}^{b-\mu}f(x)\,\mathrm dx}=:\mu+\frac{p(\mu)}{q(\mu)},$$ and $$m'=1+\frac{p'q-pq'}{q^2}=\frac1q\Big(q+p'-\frac pqq'\Big),$$ where the prime indicates differentiation by $\mu$. Of course \begin{align} p'&=af(a)-bf(b),\\ q'&=f(a)-f(b), \end{align} and $p/q=m$ at $\mu=0$, in which case \begin{align} m'|_{\mu=0}&=\frac1q\Big(q+af(a)-bf(b)-m\big(f(a)-f(b)\big)\Big)\\ &=\frac1q\Big(q-(m-a)f(a)-(b-m)f(b)\Big). \end{align} (We can assume $\mu=0$ with no loss of generality; just consider a shifted $f$ instead.)

This has a nice geometric interpretation. In the figure below, $q$ is the area below the blue curve while $(m-a)f(a)+(b-m)f(b)$ is the area below the red one. So $m'$ is positive if and only if the blue area is greater than the red. I don't know how to prove it though. It's certainly not true for an arbitrary probability distribution, even assuming it is unimodal.

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Very nice and great graph. Unfortunately, I also have no idea where to go from here. (And the reddish area should be $(m-a)f(a)+(b-m)f(b)$.) – Nameless Jul 16 '13 at 21:49

If we look at $$\frac{\partial E[x|a\le x\le b]}{\partial \mu}=1+\frac{[-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2}{[\Phi(\beta)-\Phi(\alpha)]^2}$$

then $$\frac{[-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]-[\phi(\alpha)-\phi(\beta)]^2}{[\Phi(\beta)-\Phi(\alpha)]^2} \leq \frac{[-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]}{[\Phi(\beta)-\Phi(\alpha)]^2}$$

It seems that it suffices to show (or not show) that $$\left|\frac{[-\phi'(\alpha)+\phi'(\beta)][\Phi(\beta)-\Phi(\alpha)]}{[\Phi(\beta)-\Phi(\alpha)]^2} \right| < 1$$

Now $(\Phi(\beta)-\Phi(\alpha))^{2}$ could potentially be a very small number. This simplifies to $$\left|\frac{[\phi'(\beta)-\phi'(\alpha)}{[\Phi(\beta)-\Phi(\alpha)]} \right| < \infty$$

Added. According to this, there is no global maxima. So it seems that it is not monotone for $\mu$.

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Well, no global maximum might still be consistent with monotonicity. But we'd have to know whether there are local maxima to be sure. – Nameless Jul 16 '13 at 21:52