Consider the function $Y = 1 - 2 \Phi((c_j - \mu)/\sigma) + 2 \Phi^2((c_j - \mu)/\sigma)$

where $\Phi$ is the cumulative distribution function for the standard normal distribution and $c_j$ is a uniformly distributed random variable on the range -1 to 1. I would like to be able to express the expected value of $Y$ in terms of $\mu$ and $\sigma$.

It seems clear that $E(Y)$ is increasing in the distance between $E[c_j]$ and $\mu$, and decreasing as $\sigma$ increases. However, I can also see (from simulating) that the effect of $\sigma$ on $Y$ is dependent on $c_j - \mu$. This suggests that there is an interaction between $\sigma$ and $\mu$.

My question is whether there is a way of describing this relationship analytically? I'm afraid that I'm a bit useless at this, and although the simulated results might be sufficient for my purposes, I would like to be able to show this a little more concisely, and suspect I am missing something obvious. Any suggestions appreciated!

The plot below shows the results of the simulation: enter image description here

Notes: in the simulation I am treating $c_j$ as uniformly distributed, varying the values of $\sigma$ and $\mu$ and simply plugging them into the formula above before taking an average.

  • 2
    $\begingroup$ Please check your parentheses carefully in the latter half of your formula. Writing something like $\Phi(x)^2$ (where $x$ is $(c_j-\mu)/\sigma)$) is ambiguous: did you mean $[\Phi(x)]^2$ (which could also be written as $\Phi^2(x)$ in analogy with $\sin^2(x)$ denoting the square of the number that is the sine of $x$? or was it a typo for $\Phi\left(x^2\right)$? $\endgroup$ – Dilip Sarwate Dec 22 '15 at 23:22
  • 1
    $\begingroup$ Apologies, I've clarified the equation according to your suggestion - it was supposed to denote $[\Phi(x)]^2$, though I have adopted your suggested notation. Thanks. $\endgroup$ – user2728808 Dec 22 '15 at 23:37

As $C$ varies from $-1$ to $+1$, the function $\Phi\left(\frac{C-\mu}{\sigma}\right)$ is a slowly increasing function whose value increases from varies $\Phi\left(\frac{-1-\mu}{\sigma}\right)$ to $\Phi\left(\frac{1-\mu}{\sigma}\right)$.

A more generic question is:

What is $E[\Phi(X)]$ when $X$ is uniformly distributed on (a,b)?

The answer can be obtained via integration by parts and use of the result $\frac{\mathrm d}{\mathrm dx}\phi(x) = -x\cdot \phi(x)$ where $\phi(x)$ is the standard normal density function. We have that \begin{align} E[\Phi(X)] &= \frac{1}{b-a}\int_a^b \Phi(x)\,\mathrm dx\\ &= \left.\left.\left.\frac{1}{b-a}\right[\Phi(x)\cdot x\right\vert_a^b - \int_a^b \phi(x)\cdot x \,\mathrm dx\right]\\ &= \left. \frac{b\Phi(b) - a\Phi(a)}{b-a} + \frac{\phi(x)}{b-a}\right\vert_a^b\\ &= \frac{b\Phi(b) - a\Phi(a) + \phi(b)-\phi(a)}{b-a}. \end{align}

Similarly, \begin{align} E\left[\Phi^2(x)\right] &= \frac{1}{b-a}\int_a^b \Phi^2(x)\,\mathrm dx\\ &= \left.\left.\left.\frac{1}{b-a}\right[\Phi^2(x)\cdot x\right\vert_a^b - \int_a^b 2\Phi(x)\phi(x)\cdot x \,\mathrm dx\right]\\ &= \frac{b\Phi^2(b) - a\Phi^2(a)}{b-a} - \frac{1}{b-a}\int_a^b 2\Phi(x)\phi(x)\cdot x \,\mathrm dx\\ &= \frac{b\Phi^2(b) - a\Phi^2(a)}{b-a} - \left.\left.\left.\frac{2}{b-a}\right[-\Phi(x)\phi(x)\right\vert_a^b + \int_a^b \phi^2(x)\,\mathrm dx\right]\\ &= \frac{\left(b\Phi^2(b) - a\Phi^2(a)\right)+2\left(\Phi(b)\phi(b) \right)-2\left(\Phi(a)\phi(a)\right)}{b-a}\\ &\qquad\qquad- \frac{1}{(b-a)\sqrt{\pi}}\int_a^b \frac{e^{-x^2}}{\sqrt{\pi}}\, \mathrm dx \end{align} Now, $\displaystyle \frac{e^{-x^2}}{\sqrt{\pi}}$ is the density of a normal random variable $Z$ with mean $0$ and variance $\frac 12$, and so that last integral is just $P\{a < Z < b\} = \Phi\left(\sqrt{2}b\right)-\Phi\left(\sqrt{2}a\right)$. I will leave to you the task of working out the details, then plugging in $\frac{\pm 1 - \mu}{\sigma}$ for $b$ and $a$ in the above formulas and finally figuring out $E[Y]$.

  • $\begingroup$ This is great, thanks so much for the help. I am still working through this (I'm new to calculus!) but was wondering if you could explain a little more what you mean by "Now, $\displaystyle \phi^2(x) = \frac{e^{-x^2}}{2\pi}$ is just $\frac{1}{\sqrt{\pi}}$ times the density of a $N(0,\frac 12)$ random variable and so $\int_a^b \phi^2(x)\,\mathrm dx$ can be evaluated in terms of $\Phi(\cdot)$." Apologies if I'm missing something obvious, but I don't quite follow. $\endgroup$ – user2728808 Dec 23 '15 at 10:55
  • $\begingroup$ @user2728808 See revised version. $\endgroup$ – Dilip Sarwate Dec 23 '15 at 13:27

Random variable $Y$ can be expressed as:

enter image description here

where Erf[z] denotes the error function $\frac{2}{\sqrt{\pi }}\int _0^z e^{-t^2}d t$, and where $X \sim \text{Uniform}(-1,1)$ with pdf $f(x)$:

enter image description here

Then, $E[Y]$ can be solved analytically as:

enter image description here

where I am using the Expect function from the mathStatica add-on to Mathematica to do the nitty-gritties.

While the result is not necessarily pretty, it is exact and symbolic (which is what the OP was seeking), and one can differentiate it, or plot it etc.

Here is a plot of the solution $E[Y]$, as $\sigma$ increases, when $\mu = 0$ (blue), $\mu = 1$ (orange), and $\mu = 2$ (green)

enter image description here


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.