Foundations of Data Science

Example: Let f (x) = x k for k an even positive integer. Then, f ′′ (x) = k(k − 1)x k−2
which since k − 2 is even is nonnegative for all x implying that f is convex. Thus,
E (x) ≤ k√ E (x k ),
since t 1 k is a monotone function of t, t > 0. It is easy to see that this inequality does not
necessarily hold when k is odd; indeed for odd k, x k is not a convex function.
Tails of Gaussian
For bounding the tails of Gaussian densities, the following inequality is useful. The
proof uses a technique useful in many contexts. For t > 0,
∫ ∞
x=t
e −x2 dx ≤ e−t2
2t .
In proof, first write: ∫ ∞
x=t e−x2 dx ≤ ∫ ∞ x
x=t t e−x2 dx, using the fact that x ≥ t in the range of
integration. The latter expression is integrable in closed form since d(e −x2 ) = (−2x)e −x2
yielding the claimed bound.
A similar technique yields an upper bound on
∫ 1
x=β
(1 − x 2 ) α dx,
for β ∈ [0, 1] and α > 0. Just use (1 − x 2 ) α ≤ x β (1 − x2 ) α over the range and integrate in
closed form the last expression.
∫ 1
x=β
(1 − x 2 ) α dx ≤
∫ 1
x=β
= (1 − β2 ) α+1
2β(α + 1)
x
β (1 − x2 ) α dx =
∣
−1
∣∣∣
1
2β(α + 1) (1 − x2 ) α+1
x=β
12.4 Probability
Consider an experiment such as flipping a coin whose outcome is determined by chance.
To talk about the outcome of a particular experiment, we introduce the notion of a random
variable whose value is the outcome of the experiment. The set of possible outcomes
is called the sample space. If the sample space is finite, we can assign a probability of
occurrence to each outcome. In some situations where the sample space is infinite, we can
assign a probability of occurrence. The probability p (i) = 6 1
for i an integer greater
π 2 i 2
than or equal to one is such an example. The function assigning the probabilities is called
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