In part 1, we defined the term strictly concave and introduced the theorem that if $f$ is strictly concave and $f(x)=f(x+1)$, then the integer(s) that maximize $f$ are exactly the set $\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\}$ which is usually a set containing only one number $\mathrm{round}(x+1/2)$.  In part 2, we showed how the theorem in part 1 can be used to prove that the maximum value that can be obtained when putting an integer into $f(x)=a x^2 +  b x + c$ is $f(i)$ where $i=\mathrm{round}(-b/(2a))$ (assuming $a<0$).

The next theorem says that the real number $x$ that maximizes $f$ is always near the integer $i$ that maximizes $f$.



  • $f$ is a real valued strictly concave function,
  • there is a real number $x$ such that $f(x)$ is the maximum value that $f$ attains with real inputs, and
  • there is an integer $i$ such that $f(i)$ is the maximum value that $f$ attains with integer inputs,

then $|x-i|<1$.

The idea of the proof is fairly simple.  Suppose that $i>x+1$.  Then $i-1>x$. So $i-1$ is between $i$ and $x$.  Concavity implies that that the point $(i-1, f(i-1))$ lies above the line connecting $(x, f(x))$ and $(i,f(i))$, but $f(x)\geq f(i)$.





The fact that the point $(i-1, f(i-1))$ is above the line implies that $f(i-1)> f(i)$, but that contradicts the definition of $i$, so using proof by contradiction, $i\leq x+1$.  The other cases are either similar or easy.


Find the maximum value of  $f(x) = x (x-14) (x-20)$ among all integers between 0 and 14.


Finding the maximum among reals is easy if you know differential calculus and the quadratic formula.  According to differential calculus, the maximum can only occur if the “derivative” of $f(x)$ is zero.   $$f(x) = x (x-14) (x-20) =x^3-34 x^2 + 280x.$$ The derivative of a polynomial can be found by multiplying the coefficient by the exponent and decreasing the exponent by 1. The result is $$f'(x) = 3 x^2 -68 x + 280.$$ If we put that in the quadratic formula, we get $$x=\frac{68\pm\sqrt{68^2 – 4 \cdot 3 \cdot 280}}{2 \cdot 3} = \frac{68\pm\sqrt{1264}}{6}\approx 5.41\ \mathrm{or\ } 17.26.$$ It turns out that 17.26 is a local minimum, so the local maximum occurs when $x\approx 5.41$.  According to the theorem, any integer $i$ that maximizes $f(i)$ must be within 1 of 5.41, so that integer can only be 5 or 6.  Now $f(5)= 675$ and $f(6) = 672$, so 5 is the integer that maximizes $f$.  Incidentally, $f(5.41) = 678.025021$ and the true maximum of $f(x)$ is $$f\left(\frac{68-\sqrt{1264}}{6}\right) = \frac{16}{27} \left(442+79 \sqrt{79}\right)\approx 678.025101610626.$$

We could actually use the theorem from part 1 to find the integer $i$ that maximizes $f(i)$.  Solving $f(x)=f(x+1)$ using the quadratic formula gives $$x=\frac{1}{6} \left(65\pm\sqrt{1261}\right).$$ If you replace the $\pm$ with the plus sign, you get the local minimum.  If you replace it with the minus sign, you get $$\frac{1}{6} \left(65-\sqrt{1261}\right)\approx 4.91.$$ According to the theorem from part 1, the integer that maximizes $f(x)$ must be round(4.91+1/2)=round(5.41)=5.  Often it is easier to find the maximum of $f$ rather than trying to solve $f(x)=f(x+1)$, so that is why the theorem above is useful.

If you want to see a more formal mathematical writeup, click here.  That PDF shows how you can generalize the three given theorems to a larger class of functions that are not necessarily strictly concave.  Also, in the last section of the paper, we sharpen the bound on the difference between the real maximum and the integer maximum.  Above, we showed that this difference was less than 1.  If $f$ is continuously twice differentiable, then sharper bounds can be computed.

This concludes the three part series.


In part 1, we defined the term strictly concave and introduced the theorem that if $f$ is strictly concave and $f(x)=f(x+1)$, then the integer(s) that maximize $f$ are exactly the set $\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\}$ which is usually a set containing only one number $\mathrm{round}(x+1/2)$.


The simplest concave functions are parabolas with the vertex at the top.  All parabolas can be described by three real numbers $a$, $b$, and $c$, and the formula $y=a x^2 + b x + c$.  For strictly concave parabolas $a<0$.  For example, if $a=-2$, $b=3$, and $c=5$, then the function $y= f(x) = -2x^2 + 3 x +5$ is the parabola shown below.



If the vertex (shown above in blue) is above the x-axis, then the x-coordiante of the vertex can be computed by finding the points where the concave parabola crosses the x-axis (shown above in orange).  In our case, the parabola $y=f(x)$ crosses the x-axis when $0=y=f(x)$, or when $$\begin{aligned}0&= -2x^2 + 3 x +5\\0&=-(2x^2-3×-5)\\0&=-(2×-5)(x+1).\end{aligned}.$$ If $0=\alpha\cdot \beta$, then either $\alpha=0$ or $\beta=0$.  So, the parabola crosses the x-axis when either $0=2×-5$ or $0=x+1$ which means that it crosses when $x=5/2$ or $x=-1$.  Parabolas are symmetric about the vertical line going through the vertex, so the x-coordinate of the vertex is half way between the x-coordinates of the crossing points (a.k.a the roots of $f(x)$).  $$\begin{aligned} x_\mathrm{vertex} &= \frac{x_\mathrm{crossing1} + x_\mathrm{crossing2}}2\\&= \frac{ -1 +5/2}2\\&=\frac{3/2}2\\&=3/4.\end{aligned}$$

So the x-coordinate of the vertex is $x=3/4$.  This is also the real number that maximizes $f(w)$ over all real numbers $w$.

In general, the x-coordinates of the crossing points can be found with the quadratic formula  $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2 a}.$$ The  $\pm$ sign means that one crossing can be found by replacing $\pm$ with + and the other can be found by replacing the $\pm$ with -.  But, if you have one number that is $x_1=\alpha+\beta$ and another that is $x_2=\alpha-\beta$, then the average of the two numbers is just $\alpha$.  In other words if the two values of $x$ are $x=\alpha\pm\beta$, then the average value is just $\alpha$.  So, to computer the average x-coordinate of the crossing points, all we have to do is remove the $\pm$ sign and whatever it is applied to from the quadratic formula.  $$ x_\mathrm{vertex} = \frac{-b}{2 a}.$$

theorem #2

Informally, if $y=f(x)$ is a parabola, $f(x) = a x^2 + b x + c$, and $a<0$, then the integer(s) that maximize $f(x)$ are the set $\{\mathrm{floor}(x+1/2), \mathrm{ceil}(x-1/2)\}$ where $x=\frac{-b}{2 a}$.

If $x$ is an integer plus 1/2 (e.g. 2.5, 7.5, …), this set has two elements $x+1/2$ and $x-1/2$.  If $x$ is not an integer plus 1/2, the set has only one element $\mathrm{round}(x)$ and that is the integer that produces the highest possible value of $f(z)$ among all integers $z$.


Example 1: If $f(x)= -2x^2 + 3 x +5$, then $a=-2$, $b=3$, and $c=5$.  The values of $x$ in the theorem is the same as the x-coordinate of the vertex $$x=\frac{-b}{2a} = \frac{-3}{2\cdot(-2)}= \frac{-3}{-4}=3/4.$$ The integer(s) that maximize $f(z)$ among all integers $z$ are the set  $$\begin{aligned}\{\mathrm{floor}(x+1/2), \mathrm{ceil}(x-1/2)\}&= \{\mathrm{floor}(3/4+1/2), \mathrm{ceil}(3/4-1/2)\}\\&= \{\mathrm{floor}(5/4), \mathrm{ceil}(1/4)\}\\&=\{1\}.\end{aligned}$$


Example 2: If we lower the parabola from example 1 by a little bit setting $f(x)= -2x^2 + 3 x +\sqrt{17}$, then $a=-2$, $b=3$, and $c=\sqrt{17}$.  The value of $x$ in the theorem is the same as the x-coordinate of the vertex $x=\frac{-b}{2a} =3/4.$  The result does not depend on $c$, so as in example 1, the integer that maximizes $f(z)$ among all integers $z$ is $z=\mathrm{round}(3/4)=1$.


Example 3: $f(x)= -x^2+x = (1-x)x$, then $a=-1$, $b=1$, and $c=0$.  The value of $x$ in the theorem is the same as the x-coordinate of the vertex $$x=\frac{-b}{2a} =\frac{-1}{2\cdot (-1)}=1/2.$$ The integer(s) that maximize $f(z)$ among all integers $z$ are the set  $$\begin{aligned}\{\mathrm{floor}(x+1/2), \mathrm{ceil}(x-1/2)\}&= \{\mathrm{floor}(1/2+1/2), \mathrm{ceil}(1/2-1/2)\}\\&= \{\mathrm{floor}(1), \mathrm{ceil}(0)\}\\&=\{1,0\}.\end{aligned}$$ The integers that maximize $f(z)$ among all integers $z$ are 0 and 1.  If we look at the graph of $f(x)$ below, we can see that the graph crosses the x-axis at $x=0$ and $x=1$.  So, $f(0)=f(1)=0$.  All other integers inputs produce negative values.



the spirit of the proof

It turns out that Theorem 2 can be proven from theorem 1.  Recall that in Theorem 1, we wanted to find the value of $x$ where $f(x)=f(x+1)$.  If we found that value, then the integer(s) that maximize $f$ are exactly the set $\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\}$.  In Theorem 2, $f(x) = a x^2 + b x + 1$.  So if $f(x)=f(x+1)$, then $$\begin{aligned}a x^2 + b x + c &= a (x+1)^2 + b (x+1) + c\\a x^2 + b x &= a (x+1)^2 + b (x+1)\\  a x^2 + b x &= a (x^2+2x+1)+ b x+b\\   a x^2  &= a x^2+2ax+a+ b \\    0  &= 2ax+a+ b \\ -a-b  &= 2ax \\ \frac{-a-b}{2a}  &= x\\ \frac{-b}{2a}-\frac12  &= x\\\end{aligned}.$$

Thus the integers that maximize $f(x)$ must be $$\begin{aligned} \{\mathrm{floor}(x+1), \mathrm{ceil}(x)\ \}&= \{\mathrm{floor}(\frac{-b}{2a}-\frac12+1), \mathrm{ceil}(\frac{-b}{2a}-\frac12)\} \\ &= \{\mathrm{floor}(\frac{-b}{2a}+\frac12), \mathrm{ceil}(\frac{-b}{2a}-\frac12)\}.\end{aligned}$$


Why DId I WANTED to know this

I was looking at several games where the optimal strategy depended on finding an integer $t$ that maximized $f(t)$.  In the game Slay the Spire, I wanted to maximize the amount of poison damage that I was going to do with the cards “Noxious Fumes” and “Catalyst”.  If I played the “Catalyst” on turn $t$ and the combat ended on turn $T$, then the poison damage done was $$\frac{(t+1)t}{2} -1 + (T-t)t$$ where  $ \frac{(t+1)t}{2}-1$ (note the triangle number) was the damage done by  “Noxious Fumes” and $ (T-t)t$ was the additional damage done by playing the Catalyst.  I wanted to maximize $f(t) = (T-t)t = -t^2 + T t$.

Using Theorem 2, $a=-1$, $b=T$, and $c=0$.  The Theorem says that the maximum damage occurs when you play the “Catalyst” on round $t$ where $t$ is contained in the set $\{\mathrm{floor}(x+1/2), \mathrm{ceil}(x-1/2)\}$ with $x=\frac{-b}{2 a}$.  So $x=\frac{-T}{2\cdot(-1)}=T/2$.  The best time to play the Catalyst was around half way through the combat. The catalyst should be played on round $T/2$ if $T$ is even.  If $T$ is odd, then the best round to play the “Catalyst” was for $t$ in the set $$\{\mathrm{floor}(T/2+1/2), \mathrm{ceil}(T/2-1/2)\}= \{\frac{T+1}{2}, \frac{T-1}{2} \}.$$



So the first two theorems formalized rules of thumb that I though were true.

  1. If you can find an $x$ where $f(x)=f(x+1)$, then the optimal integer(s) is $z=\mathrm{round}(x+1/2)$ with a special round that provides two answers if $x$ is an integer, and
  2. If $f(x)=a x^2 + b x + c$ (a parabola), then the optimal integer(s) is $z=\mathrm{round}(x)$ where $x=\frac{-b}{2 a}$ which is the $x$ that maximizes $f(w)$ over all real numbers $w$ (i.e. the x-coordinate of the vertex).



If you want to see a more formal mathematical writeup, click here.

I proved a few simple but interesting theorems about integers that maximize a function $f$ when $f$ is a strictly concave function that maps real numbers to real numbers.  For example, the real number that maximizes $f(x)=x(1-x)$ is $x=1/2$, but among all the integers, the maximum possible value of the function is 0. And that maximum is achieved twice with integers 0 and 1, $f(0)=0=f(1)$.



A function is strictly concave if you can pick any two points on its graph, draw a line between them, and the curve $y=f(x)$ between the two points lies entirely above the line between the two points.



Theorem #1

Informally stated, the first theorem is that if you can find a real number $x$ such that $f(x)=f(x+1)$, then the integer(s) that maximize $f$ are the set $\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\}$ where “floor” just means round down and “ceil” means round up.  That set usually contains only one element which is $\mathrm{round}(x+1/2)$, but if $x$ is an integer, then it will contain two consecutive integers $x$ and $x+1$.

For example, if $f(x) = 4-4x^2$, then the value of $x$ that satisfies $f(x)=f(x+1)$ is $x=-1/2$ because $f(-1/2)=3=f(1/2)$.  So the theorem says that any integer that maximizes $f$ must be in the set  $$\begin{aligned}\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\} &= \{\mathrm{floor}(-1/2+1), \mathrm{ceil}(-1/2)\}\\&= \{\mathrm{floor}(1/2), \mathrm{ceil}(-1/2)\} \\&= \{0\}.\end{aligned}$$ The integer that does the best is 0 which is also the real number that maximizes $f$.



Here is another example.  If $f(x) = \sin(x)$, then the value of $x$ that satisfies $f(x)=f(x+1)$ is $x=1.0708$ because $f(1.0708)=0.877583=f(2.0708)$.  So the theorem says that any integer that maximizes $f$ must be in the set  $$\begin{aligned}\{\mathrm{floor}(x+1), \mathrm{ceil}(x)\} &= \{\mathrm{floor}(1.0708+1), \mathrm{ceil}(1.0708)\}\\&= \{\mathrm{floor}(2.0708), \mathrm{ceil}(1.0708)\} \\&= \{2\}.\end{aligned}.$$  The integer that does the best is 2.



So, that was the first theorem.  In part 2, we state a second theorem about finding the integer which maximizes a quadratic with some examples and one application the game “Slay the Spire”.

If you want to see a more formal mathematical writeup, click here.

So, I have played about 300 hours of Slay the Spire since I got it on July 26.  It’s a turn-based deck building game.  Many of these deck building games have interesting mathematics, so I have been devoting a fair amount of time to analyzing the game and writing about the game.

The most interesting theorem about the game is

D = h (( d – c b + w)/a + b/alpha)

where D is the total damage that you take, h is the total amount of damage that you give to the enemy, d is the average attack per turn from the enemy, c is the average number of cards you play per turn, b is the average block per blocking card played, w is the average amount of waisted block per turn, and alpha is the average attack for the attack cards you played.  (PDF slides here.)

The nice thing about the formula is that h, d, c, b, and alpha are often precisely known or easy to calculate.  Also, a large portion of the reasonable strategies have w=0.  If you know h,d,c,b, and w and they are constant, then the correct strategy is simple:  if (d-c b + w) is positive, then don’t block.  If it’s negative, then block a lot.

There are other analysis techniques that were mentioned in the Sept 27, 2020 reddit post “Simplified Spire Puzzles“.  My favorite is looking at the ratio of damage received to damage given.

Also, I wrote a computer program that computes the best possible strategy for just about any Slay Spire combat situation.  The drawback is that if you have over 20 cards and 10 different types of cards, the program needs about 10 terabytes of ram and 3 years of cpu time to computer the answer.  It is much quicker if you have only 10 cards the are all strike or block cards in which case it takes less than one cpu second and only a few kilobytes of ram.

I have been wondering how to present this information to the world.  Today, I showed the formula and my program to my friends Cat and Chuck who are both a) fans of Slay the Spire, and b) programmers.  Additionally, I created about 10 power point slides and a 16 page document mostly describing simplified Slay the Spire problems and their solutions.

Additionally, I would like to document all the card combinations that result in either an infinite sequence of actions (resulting from hyperbolic growth) or another kind of growth.  Growth in this game seems to be limited to quadratic (1 4 9 16 25…), cubic (1 8 27 64 125…), or exponential (1 2 4 8 16 32…).  I have never seen any other kind of growth in any game except dominion which  can have polynomial growth where the exponent is a fraction.

I don’t mind writing, but I do mind rewriting and combining my short essays into a larger, more useful work especially if no one is going to read it.

Cheers, Hein


I’ve been reading “Category Theory for Programmers” which was suggested to me by Mark Ettinger.  This book presents many examples in C++ and Haskell.  It teaches you some Haskell as you read the book.  It uses almost zero upper level mathematics and it skips almost all of the mathematical formalities.  If you decide that you want to read it, then you might want to read the first six chapters of “Learn You a Haskell for Great Good!” and write a few small Haskell programs first.  (I also would suggest trying to solve the first three problems in Project Euler  using Haskell.)

I find the book to be easy to read and informative.  When the author makes a statement like   A*(B+C) = A*B + A*C where * means categorical product, + means coproduct, and = means isomorphic, I find myself trying to figure out the categories where the statement is true and the categories for which it is false.  (It is true for the Category of Set and the Category Hask.  The book is mostly about those categories.) That type of thinking improves my understanding of category theory.  The book is also reawakening the parts of my brain that had forgotten parts of category theory and Haskell.

Interestingly, in category theory, $A*(B+C) = A*B + A*C$ can be translated into the following theorems :

  1.  A*(B+C) = A*B + A*C  is true for all positive integers A,B, and C,
  2. max(A, min(B,C)) = min( max(A,B), max(A,C))  for all real numbers A, B, and C,
  3. lcm(A, gcd(B,C)) = gcd( lcm(A,B), lcm(A,C) )   where  lcm means least common multiple and gcd means greatest common denominator, and
  4. intersection(A, union(B,C)) = union( intersection(A,B), intersection(A, C)).
If you don’t believe the four theorems, here is some Mathematica Code which tests each theorem:
test[ funcRandChoose_, prod_, sum_, i_] := Tally[ Table[ 
     A = funcRandChoose[];
     B = funcRandChoose[];
     C = funcRandChoose[];
      prod[ A, sum[B, C]] == sum[ prod[A, B] , prod[A, C]], 

test[ RandomInteger[{1, 1000}] &, Times, Plus, 100]
test[ RandomInteger[{-1000, 1000}] &, Max, Min, 100]
test[ RandomInteger[{-1000, 1000}] &, LCM, GCD, 100]
test[ RandomSample[ Subsets[ Range[5]]] &, Intersection, Union, 100]

=\sum_{n=0}^\infty \frac{x^n}{n!}\cdot\sum_{n=0}^\infty \frac{(-x)^n}{n!}
=\sum_{n=0}^\infty \frac{x^n}{n!}\cdot\sum_{n=0}^\infty(-1)^n \frac{x^n}{n!}
=\sum_{n=0}^\infty\sum_{i=0}^n (-1)^i \frac1{i!}\frac1{(n-i)!}x^n \quad\mathrm{collecting\ coef\ of\ } x^n
= 1+ \sum_{n=1}^\infty \frac{ (1-1)^n}{ n!} x^n = 1
using the binomial theorem on the second to last equality.

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 $0< p<1$.)

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