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I am viewing a video from MIT OpenCourseWare. The course name is Discrete Stochastic Processes, Course number 6.262.

In the second video lecture, at minute 46:53, the ratio of the same Binomial Random Variable was explored. The purpose is to show the probability density function of the Binomial Random Variable increases and decreases for certain values of the input. I agree with this, from this specific derivation. However, I am curious about some of the specific mathematical statements.

From the attached image, Screenshot at 46:53,

it can be seen that $$ \frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} = \frac{n-k}{k+1} \frac{p}{q}\ldots (\text{1})$$

$p$ and $q = 1-p$ are real values in the set $[0,1]$. $n$ is fixed, and $0 \leq k \leq n$. It appears useful to set the ratio in equation (1) either greater than or less than 1, to show the increasing / decreasing nature of $\mathrm{P}(k)$. So doing this, I get:

$$ \frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} < 1 \text{ when } k > np-q$$ $$ \frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} > 1 \text{ when } k < np-q$$

It is at this point that I am unable to verify the statements in the slides. How do I reach the middle part, with the three cases? My cases don't exactly match up with the cases on the slides.

When the ratio $\frac{f(k+1)}{f(k)}$ is greater than 1 and function $f$ is non-negative, it means the value of the function at 1 unit ahead is greater than the function at the current position. It could be the case that the function is increasing, yet the ratio is showing a value less than 1 (if we're nearing the top of the hill).

How was the chart plotted? I see the notation for the floor of the value $pn$, but I was unable to reach that.

Lastly, how was the final statement reached? I think the floor function is part of the difficulty, and the general quantities $q$, $p$, which are between 0 and 1 (which would tend to agree with use of the floor function).


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If $np-q = np-(1-p) = (n+1)p-1$ is not an integer, then the sequence of values $$P(0), P(1), \ldots, P(n)$$ has a unique maximum. If $np-q$ is an integer, then the maximum is not unique, and the sequence increases, stays constant for two values, and decreases beyond that. In other words, two successive values in the sequence are equal to the maximum. –  Dilip Sarwate Aug 29 '12 at 13:13
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1 Answer

Your statement $\frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} < 1 \text{ when } k > np-q$ implies (when combined with $q\ge 0$) $$\frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} < 1 \text{ when } np <k .$$

Your statement $\frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} > 1 \text{ when } k < np-q$ implies (when combined with $q\le 1$) $$\frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} > 1 \text{ when } np > k+1 .$$

The middle expression is simply saying that the change happens somewhere in the gap in-between and that $\frac{\mathrm{P}(k+1)}{\mathrm{P}(k)} $ is close to $1$ here.

The graph with $x$-axis $k-\lfloor pn\rfloor$ is simply saying the peak of $\mathrm{P}(k)$ happens at $x=0$ or $x=1$.

The "In other words..." looks like a deliberate looseness of language and might end better with "...decreasing for $k \gt \lceil pn \rceil $.

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I am curious on how $A - B < C \Rightarrow A < C$, when $B > 0$. If I set $A = 4, B = 2, C = 3$, then the implication doesn't seem to be true. However, the reverse direction $A < C \Rightarrow A - B < C$ does hold in general for any $A,B,C$. –  jrand Aug 29 '12 at 15:08
@jrand: It is the reverse that matters: if proposition $P$ is true when $A−B<C$ for $B$ positive then this implies proposition $P$ is true when $A<C$. If you like: $(P \Leftarrow Q) \& (R \Rightarrow Q) \implies (P \Leftarrow R)$ –  Henry Aug 29 '12 at 16:08
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