Transcript Lecture 14w Frequency Analysis

Chapter Three
Frequency Analysis
To lessen flood
damage, we need to
know the magnitude
and frequency of the
floods a watershed is
likely to get.
Notice here, Houston gets 12
inches of rain in a day every 100
years., ranging upwards from
0.5 in/hr for 24 hrs. Thus for
Houston, the 100 year storm is
12 inches deep.
Most of the measurements we make in
Hydrology (flow rate, rainfall intensity)
cannot be predicted with certainty. They are
Random Variables.
Method of average
rectangles
Most are continuous, as shown by the
smooth curve in the upper figure, but we
generally make them discrete by averaging
over a time interval. In the upper figure, the
quantized average flow between hours 15
and 20 was about 3 cfs.
Rainfall forecasts are
given as probabilities.
Note % on Precip.
The Silenz River watershed is unaltered, and the 5-year running averages
(average of current year and previous 4) is similar over 75 years.
By contrast, this example shows decreased variability after a dam
created a reservoir in the 60’s.
Part 1: Probability
• The probability P(A) of an event A is the
relative number of occurrences of event A
after a very large number of trials.
• It will be convenient to define the sum of the
probabilities of all the possibilities as one.
• Si (P(Xi) = 1
• This means that a probability can range from
zero to one
• 0≤P≤1
Independence
• Two trials are independent if the probability of
one outcome does not depend on any
previous outcomes.
• For example, in a coin toss, the probability of
tails does not depend on any previous tosses.
Unions
Eight of the 52
cards satisfy the
occurrence of
this event
• The probability of a Union “A OR B”
(occurrence of either of two mutually
exclusive events) is the sum of the
probabilities of either.
• For example, in the drawing of a single card
from a fresh deck, the probability of drawing
an Ace or a King is 4/52 + 4/52
Mutually exclusive,
cannot draw a card
that is both a King and
an Ace.
OR
Intersections
• The probability of an intersection “A AND B”
(occurrence of both of two events) is the product of
the probabilities of each.
• For example, in the drawing of a single card from a
fresh deck, the probability of drawing a face card
(J,Q,K) that is red is 12/52 x 26/52 = 312/2704 = 6/52
•
There are 12 face cards (red or black)
and 26 red cards in a 52 card deck.
Intersections 2
• The probability of throwing a five with a single
die is 1/6
• The probability of repeating the five is
• 1/6 x 1/6 = 1 / 36
• The probability of throwing n fives in a row is
(1/6)n
• In n coin tosses, the chance of getting n tails in
a row is 1/2n
The T-year Flood
• We talked about an event, the T-year flood.
• For example, we talked about a 100-year flood
in Houston
• The probability of the event T is 1/T
• For example, in any year, the chance of getting
the 100-year flood is 1/100
• The probability that the event T does not
occur is 1- 1/T
Annual Maximum Event
• An annual maximum event has a return period
(or “recurrence interval”) of T years if its
magnitude is exceeded, on average, every T
years.
Annual Exceedances
• The n largest events (flood, rainfall) in n years
• May have more than one event in the same
year
• Still assumes independent
An Example: n = 3 years
• Year
1
2
3
Rank
1
2
3
three highest flows (cfs)
1000 950
550
550
400
200
900
800
100
A. Maxima
1000
900
550
A. Exceedances
1000
950
900
For the three
years each has a
maximum, the
largest in year 1,
then year three,
then year 2
But year 1 also
had the second
highest flow in
the three years,
so the three
largest
exceedances are
1000 and 950
(year 1) and 900
(year 3)
The large peaks in excess of 2800 mm are each due to a single flood. Rain
exceeded 3000 mm intervals four times, in1904, then not till1929, then 1953, then
2000
The average interval is: ( 25 + 24 + 47) / 3 = 32 year event, the 32 year flood exceeds
3000 mm.
A 200-year precipitation index for the central English Lake District
P. A. BARKER, R. L. WILBY & J. BORROWS
Hydrological Sciences–Journal–des Sciences Hydrologiques, 49(5)
October 2004 pages 769-785
Two Possible Events
• Frequently we deal with situations where the
are two possible events
• Heads or Tails
• It rains or it doesn’t
• The dam is overtopped or it isn’t
• Suppose the probability that it will rain today
is 75%, or Prain = 3/4
• Then the probability that it WON’T rain is
• Pdry = 1 – 3/4 = 1/4 or 25%
since the total has to be one.
Coin Tosses
• Suppose we have a fair coin, so that we expect
to get nearly the same number of heads and
tails if we flip the coin many times. We expect
to get heads half the time, and tails half the
time. We say that the probability of getting
heads p=1/2, and the probability of getting
tails is q =1/2 also. This is our null hypothesis,
H0., which says that the coin is fair, heads and
tails are equally likely.
Coin Tosses
Heads or tails are the only two
possibilities, i.e. we will get one
or the other each coin toss.
Each is equally likely; each
should happen 1/2 the time.
Notice 1/2 + 1/2 =1; we already
agreed that it will be convenient
if the sum of all probabilities
adds to one. Then if the
probability of one event p
equals any value less than one,
the probability of the other
event, q , must equal (1-p).
Coin Tosses
• We can do a lot of checking ideas in science by
assuming that there is no difference between
two things, and then check the frequencies we
observe against our Null Hypothesis, H0. If our
observation should be very rare, maybe we
should reject H0.
Coin Tosses
• If we flip a coin only four times, and we get
HHHH, we shrug and say we got that by
chance; heads and tails are still equally likely.
Expected values were not found because of
“sampling error”.
Coin Tosses
• But if we get
HHHHHHHHHHHHHHHHHHHHHHHT in 24
coin tosses we begin to think that our null
hypothesis is wrong, i.e. that maybe there is a
difference in the probability of getting a head
and tail.
Binomial Distribution
• We can actually calculate the chance of getting k=23
heads in n=24 coin tosses if our null hypothesis H0 is
right. We feel intuitively that 23 heads and one tail
will be a rare event. Is it? Call the event "heads" a
success.
• The probability of getting exactly k successes with
prob. p in n trials is given by the binomial density:
That Binomial Coefficient
• If there is only one way to do something, then
the binomial coefficient is 1.
• For example, suppose we want event A, the
probability of a tail ONLY on the second of five
coin tosses. That is, Head Tail Head Head Head
• There is only one way to do that, so the
probability is a pure union for 1 tail and four
heads
Prob(A) = 1/24 x 1/21 = 0.03125
That Binomial Coefficient
• BUT, If there is more than one way to do
something, then the binomial coefficient is
larger than 1.
• For example, suppose we want event B, the
probability for 1 tail and four heads, with the
tail anywhere
• There are five ways to do that, put the tail in
the 1st, 2nd, 3rd, 4th, or 5th positions.
• Prob(B) = 1/2 x 1/2 + 1/2 x 1/2 + 1/2 x 1/2 + 1/2 x 1/2 + 1/2 x 1/2
4
1
4
1
4
1
• Prob (B) = 5 x 1/24 x 1/21 = 0.15625
4
1
4
1
That Binomial Coefficient
• In complex situations, it is easier to calculate
the coefficient using the formula
•
n !
(n-k)! k!
where n is the number of trials, and k is the
number of one of the two possiblities.
That Binomial Coefficient
•
•
•
•
•
•
•
•
•
Let’s say you have a project that will last n = 10 years and the probability of a rain that
will flood your project occurs, on average, every 50 years. In any year, P(flood) = 1/50,
and P(no flood in some year) = 49/50
If floods in the first (foundation pour and cure) and the seventh year (basement electrical
and HVAC install prior to waterproof enclosure) will kill your profit, there is only one
possible ordering that will cause problems with your profit:
yes no no no no no yes no no no,
and P(flood in 1st and 7th year) = 1 x (1/50)2 x (49/50)8 = .00034
However if k = 2 floods in any of the n = 10 years will kill your profit, there are many ways
to do this
n ! = 10 ! = 10x9x8x7x6x5x4x3x2x1
(n-k) ! = (10-2) ! = 8x7x6x5x4x3x2x1
2 ! = 2x1 = 2
and “10 choose 2” = 10x9/2 (Blackboard DEMO) = 45 ways to arrange 2 floods in 10
years
P(2 floods in 10 years) = 45 x (1/50)2 x (49/50)8 = .0153
Back to our Coin Toss example
1 tail in 24 tosses
The left is
• I'll calculate it for you, just this once. The
terms
on
the
lef
(because k = 23 heads, n-k = 1 tail)
side are:
• n! = 24*23*22* … *1 ;
k! = 23*22*21* … *1 ; and
n - k = 24 - 23 = 1 so (n - k)! = 1! =1
n! !
k! (n - k)!
=
(24*23*22* … *1)
= 24
(23*22*21* … *1)( 1 )
Our Coin Toss Example
• p = 1/2 and 1-p is also =1/2, so the right side is
1/2 raised to the 23rd power times 1/2.
• The whole thing is equal to
• = 24 * (1/2)23* (1/2)1 = .00000143. In other
words 23 heads and one tail will occur about
once in 700,000 attempts in repeated 24 coin
tosses. We can probably agree that that is too
rare. Our coin probably came from a magic
shop, and we should reject H0 which says that
heads and tails are equally likely.
The basis of Inferential Statistics.
Risk and Reliability
• The probability of at least one occurrence of
the T-year event in n years is called the Risk.
• The prob. of an occurrence in one year p = 1/T
• The Risk is the sum of probs. of one such
flood, two floods, three floods, …, n floods.
• Easier to calculate with 1 - P(none) = 1 – P(0)
• P(0) = P(none in n years) = (1- p)n
• Risk = 1 – P(0) = 1 - (1- p)n = 1 – (1 – 1/T)n
• Reliability is 1-Risk = (1- p)n = (1- 1/T)n
Flood Control Design
• We will use these ideas with the examples and
related homework problems.
Earthen cofferdam overtopped
Sheet Piling cofferdam section