# Calculating a specific joint probability involving sums of binomial distributions

The following might look like a simple problem - but the question has been unanswered for more than a week on math.stackexchange.com, and I have asked quite a few of the Ph.d. students at our department - hopefully someone will come up with a shamefully simple solution. The problem originates from analyzing a specific attack in relation to a specific cryptographical protocol.

Graph-theoretical interpretation of the problem:

Fix a $k \in \{0,\ldots\}$, and pick an $x \in \{0,\ldots, 2k\}$ uniformly at random.

Consider the problem of doing a walk in the graph of the type depicted here:

1. Flip a coin $x$ times. The number of times to walk along the diagonal is decided by the number of heads.
2. Flip a coin $2k+1-x$ times. The number of times to walk right is decided by the number of heads.
3. Flip a coin $2k+1-x$ times. The number of times to walk up is decided by the number of heads.

MY QUESTION: Does $g(k,x) = \frac{1}{4}(1 + \frac{x}{2k+1})$ bound the probability that we have made at most $k$ steps in both horizontal and vertical direction? (colored red in the figure)

Representation of the problem in terms of bit strings and majority function

I am analyzing the following experiment:

1. Pick an $x \in \{0,\ldots,2k\}$ uniformly at random

2. Pick $(2k+1)$-bit bitstring $b_1=(u,v_1) \in \{0,1\}^x \times \{0,1\}^{2k+1-x}$ uniformly at random

3. Pick a $(2k+1-x)$-bit bitstring $v_2 \in \{0,1\}^{2k+1-x}$ uniformly at random

What is the probability that the majority function of $b_2 = (u,v_2)$ is bigger than the majority function of $b_1 = (u,v_1)$?

Remark: The reason for picking a bit string of length $2k+1$ is for the majority function to be well-defined.

It can be analyzed as follows. Define the random variables:

• $X \sim Uniform(\{0,\ldots,2k\})$
• $Y(x) \sim Binom(x,\frac{1}{2})$
• $Z_1(x),Z_2(x) \sim Binom(2k+1-x,\frac{1}{2})$

What is: $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \geq k+1]$?

The challenge of the problem is easiest shown by fixing a specific $x$, and calculating:

$\Pr[Y(x) + Z_1(x) \leq k \wedge Y(x) + Z_2(x) \geq k+1]$

$= \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y] \Pr[Z(x) \geq k+1-y]$

$= \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y] (1 - \Pr[Z(x) \leq k-y])$

$= \Pr[Y(x) + Z(x) \leq k] -\sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$

$= \frac{1}{2} -\sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$

where we just let $Z_1=Z_2=Z$ after the dependence has been removed.

But how to go on from here?

If we let

• $f(k,x) = \sum_{y=0}^x \Pr[Y(x) = y] \Pr[Z(x) \leq k-y]^2$
• $g(k,x) = \frac{1}{4}(1 + \frac{x}{2k+1})$

Then a plot from maple suggests that $g(k,x) \geq f(k,x)$ for all values that we consider.

An explicit way of defining $f$ is: $f(x,y) = \sum_{y=0}^x p_{x,y} \left( \sum_{z=0}^{k-y} p_{2k+1-x,z} \right)^2$ where $p_{a,b} = 2^{-a}\binom{a}{b}$.

How can I show that $g$ is an upper bound to $f$? I tried all kinds of things - everything from rewriting to expressions about the variance of some complicated variable, to trying out different induction strategies. I also looked into the theory of moment generating functions. Maybe I was just not creative enough.

If successfully proven, it will result in the lower bound

$\Pr[Y(x) + Z_1(x) \leq k \wedge Y(x) + Z_2(x) \geq k+1] \geq \frac{1}{2} - \frac{1}{4}(1 + \frac{x}{2k+1})$

Taking the average over all $x \in \{0,\ldots, 2k\}$, we end up with a lower bound on the expectation of $\frac{1}{2k+1} \sum_{x=0}^{2k} (\frac{1}{2} - \frac{1}{4}(1 + \frac{x}{2k+1})) = \frac{1}{4} \frac{k+1}{2k+1} \geq \frac{1}{8}$.

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If you get no joy here, you might like to try at stats.stackexchange.com. Clearly asking at all these different sites is suboptimal. Drop a line here if it gets answered anywhere else. – David Roberts Oct 5 '12 at 9:46
I have now clarified how $(u,v_1)$ is defined from $b_1$ (splitting into a $x$-bit bitstring, and a $2k+1-x$-bit-bitstring.) I take the majority of the whole bitstring, which has odd length, but only change the last $2k+1-x$ bits. – val11 Oct 5 '12 at 11:04

Perhaps this should be a comment, but I do not have enough "street credit" on mathoverflow to post comments. In your question, the expression ($g(x,k)$) depends on $x$. But according to the description of your experiment, $x$ was chosen randomly. So you are asking if for fixed choice of $X$ this holds? If I read the question correctly, what I am really reading is "given the experiment, what is the probability that we go at most $k$ steps right and and at most $k$ steps up", and then the question about the bounding probability would help to settle that question?
Anyway I have no answer to the question on $g(x,k)$, but the question I read can, unless I am wrong, be answered simpler. Consider the following reasoning:
With probability $\frac{1}{2}$, the number of heads in step $1$ is at most $\frac{x}{2}$. (Assume $\frac{x}{2}$ is an integer).
For the going right part, we flip $2k+1-x$ coins. The expected number of heads is $k+\frac{1}{2}-\frac{x}{2}$. The probability of the number of heads being at most $k-\frac{x}{2}$ is at least $\frac{1}{2}$. Similar for the going up part, so the probability is at least $\frac{1}{8}$.
Hi Magnus. I am searching for the average value of $f(x,k)$ over all $x$. I am just looking at it for fixed $x$, if that could make things easier. – val11 Oct 5 '12 at 20:16
What I am searching for is an upper bound on $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \leq k]$. I originally had the problem of $\Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \geq k+1]$, which I showed was equal to the following difference: $\Pr[Y(X) + Z(X) \leq k] - \Pr[Y(X) + Z_1(X) \leq k \wedge Y(X) + Z_2(X) \leq k]$. The first term is trivial to compute, so I only need an upper bound on the second term. – val11 Oct 5 '12 at 20:16
As far as I can see, when applying your strategy to find an upper bound for $f(x,k)$, I get something like: $\leq \Pr[Z(X) \leq k-\frac{X}{2}]^2\Pr[Y(X) \geq \frac{X}{2}] + \Pr[Z(X) \leq k]^2 \Pr[Y(X) \leq \frac{X}{2}]$ which is not very tight (at a first glance - $\Pr[Z(X) \leq k]$ seems to get quite big).. I'll calculate things more thoroughly tomorrow - maybe this is actually good enough... – val11 Oct 5 '12 at 20:16