Articles by hundalhh

Mathematician and Father. Into games, astronomy, psychology and philosophy.

If you have two planets in circular orbits with semi-major axes $a_1$ and $a_2$, and periods $T_1$ and $T_2$, then the amount of time spent in retrograde is $$T_\mathrm{retro} = T_1\left|\frac{\cos ^{-1}\left(\frac{\sqrt{r}+1}{r+\frac{1}{\sqrt{r}}}\right)}{\pi\left(1-\frac{1}{r^{3/2}}\right)}\right|$$

where $r= \frac{a_2}{a_1}$. You can apply this formula to the planets in our solar system to get approximations for their time in retrograde as seen from Earth. (These approximations are often within 10% of the correct value, but sometimes the error is larger.)

$$\begin{array}{ccc} & \text{a in AU} & T_{\text{retro}} \\ \text{Mercury} & 0.387 & 23 \text{ days} \\ \text{Venus} & 0.723 & 42 \text{ days} \\ \text{Mars} & 1.524 & 73 \text{ days} \\ \text{Jupiter} & 5.204 & 121 \text{ days} \\ \text{Saturn} & 9.54 & 138 \text{ days} \\ \text{Uranus} & 19.19 & 152 \text{ days} \\ \text{Neptune} & 30.05 & 158 \text{ days} \\ \end{array}$$

It’s interesting to note that:

  • $\lim_{r\rightarrow\infty} T_\mathrm{retro}(r) = T_1/2$,
  • $\lim_{r\rightarrow 1} T_\mathrm{retro}(r) = T_1\cdot \frac{\sqrt{2}}{3 \pi }$, and
  • $\lim_{r\rightarrow 0} \frac{T_\mathrm{retro}(r)}{r^{3/2}T_1}=1/2$. If people are interested, I can post a derivation.

If people are interested, I can post a derivation.

On Math stack exchange, purpleostrich asked “Consider random variables A, B, and C. We know that A = B + C. We also know that A and C have an MGF. Is it the case that B must have a MGF?”

Here is my answer:

You Can’t Compute the MGF

In general, you can’t compute the MGF of $B$ if you only know the MGFs of $A$ and $C$. For example, consider two possible joint distributions of $A$ and $C$:

Case 1: P( A=0 and C=0) = 1/2 and P(A=1 and C=1)=1/2. In this case, the MGFs of A and C are $(1+\exp(t))/2$ and the MGF of B is 1.

Case 2: P( A=0 and C=1) = 1/2 and P(A=1 and C=0)=1/2. In this case, the MGFs of A and C are $(1+\exp(t))/2$ and the MGF of B is $\frac{\exp(-t)+\exp(t)}2$.

Notice that in both Case 1 and Case 2 the MGFs for $A$ and $C$ were $(1+exp(t))/2$, but the MGF for $B$ changed from Case 1 to Case 2.


You can prove the MGF exists

Although you can’t computer the MGF of $B$, you can prove that $M_B(t)$ exists for $t\in D=\frac12 (Dom(M_A)\cap (-Dom(M_C))$. Suppose $t\in D$. Then $||\exp(ta)||_1<\infty$ and $||\exp(-tc)||_1<\infty$ where $||g||_p=\left(\int\int |g(a,c)|^p\; f(a,c)\; da\; dc\right)^{1/p}$ is the $L_p$-norm of $g$ over the joint probability space and $f(a,c)$ is the joint pdf of $A$ and $C$. That implies $||\exp(ta/2)||_2 < \infty$ and $||\exp(-tc/2)||_2 < \infty$. By the Hölder’s inequality or, more specifically, Schwarz inequality, $||\exp(ta)\exp(-tc)||_1<\infty$. But, $||\exp(ta)\exp(-tc)||_1= ||\exp(t(a-c)||_1= E[\exp(tB)]=M_B(t).$ This proves that $M_B(t)$ exists for $t\in D$.


If A and C are independent

If $A$ and $C$ are independent and $B = A-C$, then it must be the case that
M_B(t) = M_A(t)\cdot M_C(-t)
whenever $t\in Dom(M_A)\cap(-Dom(M_C))$ (see e.g. Wikipedia). Here is a rough proof.

If $t\in Dom(M_A)\cap(-Dom(M_C))$, then
$$M_A(t)\cdot M_C(-t) = \int_{a=-\infty}^\infty \exp(t a) dF_A(a) \cdot \int_{c=-\infty}^\infty \exp(-t c) dF_C(c)$$
= \int_{a=-\infty}^\infty \int_{c=-\infty}^\infty \exp(t (a-c)) dF_A(a) dF_C(c)
= \int_{b=-\infty}^\infty \exp(t b) dF_B(b) = M_B(t)
where $F_A, F_B$, and $F_C$ are the cumulative distribution functions of $A, B$, and $C$ respectively.



An interesting mathematical problem came up at work today.  I had to find a formula for the standard deviation of a binomial distribution given that the random variable was not zero.  I put some notes below summarizing my results.

Removing 0 from any Distribution

Suppose that you have a random variable $X$.  What are the values of $\mu_0 := E[X | X\neq 0]$ and $\sigma_0 := \sqrt{E[ (X-\mu_0)^2| X\neq 0]}$?  After doing some algebra, I got

$$\mu_0 = \bar{X}/(1-p_0), \quad\mathrm{and}$$

$$\sigma_0 = \sqrt{ \frac{\sigma_X^2 – p_0({\bar{X}}^2+\sigma_X^2)}{\left(1-p_0\right)^2}}= \sqrt{\frac{\sigma_X^2}{1-p_0} \;-\; \frac{ p_0 \bar{X}^2}{(1-p_0)^2}}$$

where $p_0:=P(X=0)$, $\bar{X}=E[X]$, and $\sigma_X := \sqrt{E\left[\left(X-\bar{X}\right)^2\right]}\,$.

Notice that if $p_0=0$ then the right hand side reduces to $\sigma_X$.

Bernoulli Distribution

If we apply the formulas above to the Bernoulli Distribution where $X$ is either 0 or 1 and $P(X=1)=p$, then $p_0 = (1-p)$, $\bar{X}=p$, and $\sigma_X^2 = p(1-p)$, so $\mu_0 = p/(1-(1-p))=1$ and

$$\sigma_0 = \sqrt{\frac{\sigma_X^2}{1-p_0} – \frac{ p_0 \bar{X}^2}{(1-p_0)^2}}=\sqrt{\frac{p(1-p)}{p} – \frac{ (1-p)p^2}{p^2}}=0.$$

That is to be expected because if $X$ is not 0, then it must be 1.

Binomial Distribution

Anyway, I really wanted to apply these formulas to the Binomial Distribution.  For the Binomial Distribution, $p_0=(1-p)^n$, $\bar{X} = np$, and $\sigma_X = \sqrt{n p (1-p)}$.  So,

$$\mu_0 = n p/(1-(1-p)^n), \quad\mathrm{and}$$

$$\begin{align}\sigma_0&= \sqrt{  \frac{n p (1-p) – (1-p)^n(n^2p^2+n p (1-p))}{\left(1-(1-p)^n\right)^2} }\\&= \sqrt{  n p \frac{ (1-p) – (1-p)^n(np+ (1-p))}{\left(1-(1-p)^n\right)^2}.}\end{align}$$

Notice that if $n=1$ then $\mu_0=1$ and $\sigma_0=0$ which makes sense because if $n=1$ and $X\neq0$ then $X$ is always 1.  Also notice that $\lim_{n->\infty} (\mu_0 – n p) = 0$ and $\lim_{n->\infty} (\sigma_0 – \sqrt{n p (1-p)}) = 0$ which is to be expected because $\lim_{n->\infty} p_0=0$. (I am assuming $0< p<1$.)

   In March of 2016, the computer program AlphaGo defeated Lee Sedol, one of the top 10 Go players in the world, in a five game match.  Never before had a Go computer program beaten a professional Go player on the full size board.  In January of 2017, AlphaGo won 60 consecutive online Go games against many of the best Go players in the world using the online pseudonym Master.  During these games, AlphaGo (Master) played many non-traditional moves—moves that most professional Go players would have considered bad before AlphaGo appeared. These moves are changing the Go community as professional Go players adopt them into their play.

Michael Redmond, one of the highest ranked Go players in the world outside of Asia, reviews most of these games on You Tube.  I have played Go maybe 10 times in my life, but for some reason, I enjoy watching these videos and seeing how AlphGo is changing the way Go is played. Here are some links to the videos by Redmond.

Two Randomly Selected Games from the series of 60 AlphaGo games played in January 2017


Match 1 – Google DeepMind Challenge Match: Lee Sedol vs AlphaGo


The algorithms used by AlphaGo (Deep Learning, Monte Carlo Tree Search, and convolutional neural nets) are similar to the algorithms that I used at Penn State for autonomous vehicle path planning in a dynamic environment.  These algorithms are not specific to Go.  Deep Learning and Monte Carlo Tree Search can be used in any game.  Google Deep Mind has had a lot of success applying these algorithms to Atari video games where the computer learns strategy through self play.  Very similar algorithms created AlphaGo from self play and analysis of professional and amateur Go games.

I often wonder what we can learn about other board games from computers.  We will learn more about Go from AlphaGo in two weeks.  From May 23rd to 27th, AlphaGo will play against several top Go professionals at the “Future of Go Summit” conference.




“Regret Analysis of Stochastic and Nonstochastic Multi-armed Bandit Problems” by Sébastie Bubeck and Nicolò Cesa-Bianchi is available in pdf format at

The book “Deep Learning” by Ian Goodfellow, Yoshua Bengio, and Aaron Courville (associated with the Google Deep Mind Team) is available in HTML format.

(See also:  Half-precision floating-point format )

Wired has a nice article about the two most brilliant moves in the historic match between AlphaGo and Lee Sedol.

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