# Segment 30 Sanmit Narvekar

#### To Calculate

1. For a set of positive values **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \{x_i\}}**
, use Jensen's inequality to show (a) the mean of their square is never less than the square of their mean, and (b) their (arithmetic) mean is never less than their harmonic mean.

Part A

This is basically applying Jensen's inequality to the function **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = x^2}**
, which leads the statement to hold since x^2 is a convex function. Specifically, consider the graph of x^2. The mean of the squares would be like a value on a segment connecting 2 points on the parabola, and the square of the mean would be the point on the parabola in the middle of those points. Since the segment will always be higher than the point on the parabola, the statement holds.

More formally:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \left( \frac{1}{N} \sum x_i \right)^2 = f \left( \frac{1}{N} \sum x_i \right) \le \frac{1}{N} \sum f(x_i) = \frac{1}{N} \sum x_i^2 }**

Which is the proof as desired.

Part B

The harmonic mean is:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = \frac{N}{\sum\frac{1}{x_i}} }**

We define our function to be **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \frac{1}{x}}**
. Then,

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H = f \left(\frac{1}{N} \sum \frac{1}{x_i} \right)}**

By Jensen's inequality:

**Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle H \le \frac{1}{N} \sum f \left( \frac{1}{x_i} \right) = \frac{1}{N} \sum x_i}**

Which is the arithmetic mean. Thus the arithmetic mean is always larger than the harmonic mean.

2. Sharpen the argument about termination of E-M methods that was given in slide 4, as follows: Suppose that **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) \ge f(x)}**
for all **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x}**
, for some two functions **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}**
and **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g}**
. Prove that, at any local maximum **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x_m}**
of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f}**
, one of these two conditions must hold: (1) **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x_m) > f(x_m)}**
[in which case the E-M algorithm has not yet terminated], or (2) **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x_m)}**
is a local maximum of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g}**
[in which case the E-M algorithm terminates at a maximum of **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g}**
, as advertised]. You can make any reasonable assumption about continuity of the functions.

By our assumption, **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x) \ge f(x)}**
for all **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x}**
. Consider a local maximum of f at x_m. First consider the case where **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x_m) = f(x_m)}**
. If we assume that both functions are continuous, then their derivatives at x_m must be 0, since f(x_m) is a local maxima, and if g's derivative at that point were not zero, then f(x_m) would either not be continuous, or it wouldn't equal g(x_m).

The other case is that **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x_m) > f(x_m)}**
, since g has to be greater at all points. In this case, EM has not terminated, since it waits until f(x_m) = g(x_m).

#### To Think About

1. Jensen's inequality says something like "any concave function of a mixture of things is greater than the same mixture of the individual concave functions". What "mixture of things" is this idea being applied to in the proof of the E-M theorem (slide 4)?

2. So slide 4 proves that some function is less than the actual function of interest, namely **Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L(\theta)}**
. What makes this such a powerful idea?