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Chapter 1- Set theory:
𝐴 ∩ (𝐵 ∪ 𝐶) = (𝐴 ∩ 𝐵) ∪ (𝐴 ∩ 𝐶)
𝐴 ∪ (𝐵 ∩ 𝐶) = (𝐴 ∪ 𝐵) ∩ (𝐴 ∪ 𝐶)
A B
DeMorgan’s Laws:
(𝐴 ∪ 𝐵)′ = 𝐴′ ∩ 𝐵 ′
(𝐴 ∩ 𝐵)′ = 𝐴′ ∪ 𝐵 ′
𝑃(𝐵) = 𝑃(𝐵 ∩ 𝐴) + 𝑃(𝐵 ∩ 𝐴′ )
𝑃(𝐴 ∪ 𝐵) = 𝑃(𝐴) + 𝑃(𝐵) − 𝑃(𝐴 ∩ 𝐵)
𝑃(𝐴 ∪ 𝐵 ∪ 𝐶) = 𝑃(𝐴) + 𝑃(𝐵) + 𝑃(𝐶) − 𝑃(𝐴 ∩ 𝐵) − 𝑃(𝐵 ∩ 𝐶) − 𝑃(𝐶 ∩ 𝐴) + 𝑃(𝐴 ∩ 𝐵 ∩ 𝐶)
Independence:
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴)𝑃(𝐵)
Chapter 2- Counting Techniques:
Permutations:
Order matters!!
𝑃(𝑛, 𝑘) = number of ordered samples of size k from n without replacement.
𝑃(5,3) = 5 ∗ 4 ∗ 3
𝑃(10,2) = 10 ∗ 9
Combinations:
Order doesn’t matter!!
𝐶(𝑛, 𝑘) = number of unordered samples of size k from n using sampling without replacement.
5∗4∗3
10∗9
5
10
𝐶(5,3) = 3∗2∗1 = ( )
𝐶(10,2) = 2∗1 = ( )
2
3
Samples With Replacement:
N objects, sample size k
Probability= 𝑛𝑘
Distinguishable permutations:
11!
How many distinct words can you make from the word Mississippi = 4!4!2!
Chapter 3- Conditional Probability:
𝑃(𝐴|𝐵) = 1 − 𝑃(𝐴′ |𝐵)
𝑃(𝐴|𝐵) =
𝑃(𝐴∩𝐵)
𝑃(𝐵)
𝑃(𝐴 ∩ 𝐵) = 𝑃(𝐴|𝐵)𝑃(𝐵)
Bayes’ Theorem:
𝑃(𝐴)𝑃(𝐵|𝐴)
𝑃(𝐴|𝐵) =
𝑃(𝐵)
𝑃(𝐵) = 𝑃(𝐴)𝑃(𝐵|𝐴) + 𝑃(𝐴′ )𝑃(𝐵|𝐴′ )
Independence:
𝑃(𝐴|𝐵) = 𝑃(𝐴)
𝑃(𝐴 ∩ 𝐵 ∩ 𝐶) = 𝑃(𝐴)𝑃(𝐵|𝐴)𝑃(𝐶|𝐴 ∩ 𝐵)
Chapter 4- Random Variables:
Continuous random variables:
The probability of any single value is zero, since there are an infinite number of values.
Probability Density Function (pdf):
𝑓(𝑥) = 𝑃(𝑋 = 𝑥)
𝑏
𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) = ∫𝑎 𝑓(𝑥)𝑑𝑥 = 𝐹(𝑎) − 𝐹(𝑏)
∞
∫−∞ 𝑓(𝑥) = 1
Continuous Discrete Function (cdf):
𝐹(𝑋) = 𝑃(𝑋 ≤ 𝑥)
𝑥
𝑃(𝑋 ≤ 𝑥) = ∫−∞ 𝑓(𝑥)𝑑𝑥
𝑃(𝑋 ≤ 𝑥) = 𝑃(𝑋 < 𝑥)
𝐹(−∞) = 0
𝐹(∞) = 1
Converting between pdf and cdf:
𝑥
𝐹(𝑥) = ∫−∞ 𝑓(𝑥)𝑑𝑥
𝑓(𝑥) = 𝐹 ′ (𝑥)
Measures of Central Tendency:
𝑉𝑎𝑟(𝑋 + 𝑌 + 𝑐) = 𝑉𝑎𝑟(𝑋) + 𝑉𝑎𝑟(𝑌) + 2𝐶𝑜𝑣(𝑋, 𝑌)
𝑉𝑎𝑟(𝑎𝑋 − 𝑏𝑌) = 𝑎2 𝑉𝑎𝑟(𝑋) + 𝑏 2 𝑉𝑎𝑟(𝑌) − 2𝑎𝑏𝐶𝑜𝑣(𝑋, 𝑌)
𝜎(𝑎𝑋 + 𝑏) = |𝑎|𝜎(𝑋)
𝐸(𝑎𝑋 + 𝑏) = 𝑎𝐸(𝑋) + 𝑏
Approximations of discrete random variables:
When using an integral to approximate a discrete case (which can only take integer values), you
need to change the limits.
𝑏+0.5
𝑃(𝑎 ≤ 𝑋 ≤ 𝑏) ≈ 𝑃(𝑎 − 0.5 < 𝑌 < 𝑏 + 0.5) = ∫
𝑎−0.5
Chebyshev’s Inequality:
𝑓(𝑦)𝑑𝑦
X is a random variable with mean 𝜇 and variance 𝜎 2 .
1
𝑃(|𝑋 − 𝜇| ≥ 𝑘𝜎) ≤ 2
𝑘
1
The probability that X is not within 2𝜎 of the mean, is 𝑘 2 .
Percentiles:
𝑃(𝑋 ≤ 𝑥𝑝 ) = 𝐹(𝑥𝑝 ) = 𝑝
The probability that X is in the 35% percentile:
𝑃(𝑋 ≤ 𝑥0.35 ) = 𝐹(𝑥0.35 ) = 0.35 Solve for x.
Moment generating function:
𝑀𝑥 (𝑡) = 𝑝𝑒 𝑡𝑥
𝑀𝑥 (0) = 1
𝑀𝑥 ′ (0) = 𝐸(𝑋)
𝑀𝑥 ′′ (0) = 𝐸(𝑋 2 )
𝑍 = 𝑎 + 𝑏𝑥
𝑀𝑍 (𝑡) = 𝑒 𝑎𝑡 𝑀𝑥 (𝑏𝑡) = 𝑒 𝑎𝑡 ∗ 𝑒 𝑏𝑡𝑥
𝑀𝑋+𝑌 (𝑡) = 𝑀𝑥 (𝑡) ∗ 𝑀𝑌 (𝑡)
Mode:
The point where 𝑓(𝑥) reaches its maximum. Set 𝑓 ′ (𝑥) = 0 (the maximum value of a function
happens with the derivative =0). Solve for x.
Median:
-If given cdf, set 𝐹(𝑥) = 0.5, solve for x
𝑥
-If given pdf, integrate between ∫0 𝑓(𝑥)𝑑𝑥 and set answer equal to 0.5, solve for x.
-For split distributions, draw graphs or substitute in limits to decide where 0.5 falls.
Standard deviations:
“What percent of claims fall within one standard deviation of the mean?”
-Find 𝐸(𝑋), 𝜎
-Make range: [𝐸(𝑋) − 𝜎, 𝐸(𝑋) + 𝜎]
-add up the percentage of claims in that range.
Integrals:
4 |𝑥|
4
2
𝑥
𝑥
∫
𝑑𝑥 = ∫
𝑑𝑥 − ∫
𝑑𝑥
−2 10
0 10
0 10
∫ 𝑥 −1 𝑑𝑥 = ln 𝑥
Chapter 5- Discrete Distributions:
Name
Binomial
Negative
Binomial
Density
f ( x) 
 p q
f ( x) 

Geometric
Hypergeometric
Poisson
Uniform
MGF
n
k
k nk

M x t   pe t  q

n
n failures before k
successes
 1  qe t
M x t   
 p




n
n failures before the
first success
 1  qe t
M x t   
 p




c(m1 , x)c(m2 , n  x)
c(m, n)
Sample of n taken
from total of m
n  k 1
n
p q
k n
f ( x)  q p
f ( x) 
x = 0,1,2…n
e  x
f ( x) 
x!
1
f ( x) 
n
Bernoulli
f ( x)  p x q n  x
Poisson Approx.
to the Binomial
e  np (np ) x
f ( x) 
x!
 is the rate
of a rare event
x = 0,1,2…n
M x (t ) 
great : n  100, np  10
Var ( X )  npq
kq
E( X ) 
p
Var ( X ) 
kq
q
p
Var ( X ) 
q
E( X ) 
E( X ) 
t
1)
N (e t  1)
M x t   pe t  q
good : n  20, p  0.05
E ( X )  np
1
e t (e Nt  1)

Var(X)
k
m  m1  m2
M x (t )  e  (e
E(X)

nm1
m
E( X )  
E( X ) 
N 1
2
E( X )  p
p2
p2
 m  m  m  n 
Var ( X )  n 1  2 

 m  m  m  1 
Var ( X )  
N 2 1
Var ( X ) 
12
Var ( X )  pq
Chapter 6- Continuous Distributions:
Name
Uniform
Density
𝑓(𝑥) =
1
𝑏−𝑎
𝑓(𝑥) =
Beta
Weibull
Pareto
Exponential
(𝑎 + 𝑏 − 1)!
𝑥 𝑎−1 (1 − 𝑥)𝑏−1
(𝑎 − 1)! (𝑏 − 1)!
𝑓(𝑥) =
𝜏𝑥
𝑓(𝑥) =
Mx t 
CDF
𝑥
𝜏−1 −( ⁄𝜃)
𝑥−𝑎
𝑏−𝑎
𝑎≤𝑥≤𝑏
𝐹(𝑥) =
𝑎, 𝑏
0≤𝑥≤1
No set formula
𝑀𝑥 (𝑡) =
𝜏
𝜏, 𝜃
𝐹(𝑥) = 1 − 𝑒 −(
𝛼𝜃 𝛼
(𝑥 + 𝜃)𝛼+1
𝛼, 𝜃
𝜃 𝛼
𝐹(𝑥) = 1 − (
)
𝑥+𝜃
1 −𝑥⁄
𝑒 𝜇
𝜇
𝜇
𝐹(𝑥) = 1 − 𝑒
𝑥⁄ )
𝜃
−𝑥⁄
𝜇
𝑬(𝑿𝒌 ) =
Gamma
Normal
Lognormal
𝑓(𝑥) =
𝑒
(−
(𝑥−𝜇)2
)
2𝜎2
𝛼, 𝜃
(𝜇, 𝜎 )
Must be calculated
with table
of values
(𝜇, 𝜎 2 )
not equal
to mean,
variance
Must be calculated
with table
of values
2
𝜎√2𝜋
Not worth it
Not worth it
𝑏+𝑎
2
𝐸(𝑋) =
N/A
𝑎
𝐸(𝑋) =
𝑎+𝑏
N/A
1
𝐸(𝑋) = 𝜃 ∗ ( ) !
𝜏
𝜽𝒌 𝒌!
(𝜶 − 𝟏). . (𝜶 − 𝒌)
𝑀𝑥 (𝑡) =
𝐸(𝑋) =
𝜃
𝛼−1
Var X 
𝑉𝑎𝑟(𝑋) =
(𝑏 − 𝑎)2
12
𝑉𝑎𝑟(𝑋)
=
𝑎𝑏
(𝑎 + 𝑏)2 (𝑎 + 𝑏 + 1)
Not worth it
use 𝑀𝑥 (𝑡) to calculate
1
1 − 𝜇𝑡
𝐸(𝑋) = 𝜇
𝑉𝑎𝑟(𝑋) = 𝜇 2
1
(1 − 𝜃𝑡)𝛼
𝐸(𝑋) = 𝛼𝜃
𝑉𝑎𝑟(𝑋) = 𝛼𝜃 2
𝐸(𝑋) = 𝜇
𝑉𝑎𝑟(𝑋) = 𝜎 2
𝑀𝑥 (𝑡) =
−𝑥
𝑥 𝛼−1 𝑒 ⁄𝜃
𝑓(𝑥) = 𝛼
𝜃 (𝑛 − 1)!
𝑒 𝑏𝑡 − 𝑒 𝑎𝑡
𝑡(𝑏 − 𝑎)
𝜏
𝑒
𝜃𝜏
𝑓(𝑥) =
E X 
𝑀𝑥 (𝑡) = 𝑒
(𝜇𝑡+
𝜎2 𝑡 2
)
2
2
N/A
𝐸(𝑋) = 𝑒
(𝜇+12𝜎2 )
𝑉𝑎𝑟(𝑋) = 𝑒 (2𝜇+2𝜎 )
2
− 𝑒 (2𝜇+𝜎 )
Chapter 7- Normal Distribution:
The normal distribution:
Has a complicated pdf and an unsolvable cdf. Use the table of values to look up the cdf for
certain values.
The Standard Normal Distribution has 𝜇 = 0 and 𝜎 = 1.
𝑋~𝑁(𝜇, 𝜎 2 )
Evaluating the standard normal:
𝑃(𝑋 ≤ 𝑥) = Φ(x)
𝑃(𝑋 < 𝑥) = 1 − 𝑃(𝑋 > 𝑥)
𝑃(𝑋 > 𝑥) = 𝑃(𝑋 ≤ −𝑥) = Φ(−𝑥) = 1 − Φ(𝑥)
Φ(𝑥) = 1 − Φ(−𝑥)
𝑃(𝑋 ≤ −𝑥) = Φ(−𝑥) = 1 − Φ(𝑥)
𝑃(𝑋 > −𝑥) = 𝑃(𝑋 < 𝑥) = Φ(𝑥)
*Values in the table are for STANDARD normal distribution with 𝜇 = 0 and 𝜎 = 1.*
What if 𝜇 = 20 and 𝜎 = 15?
𝑋~𝑁(20,225)
𝑥−𝜇
𝑥−𝜇
𝑃 (𝑋 <
) = Φ(
)
𝜎
𝜎
12 − 20
𝑃(𝑋 < 12) = 𝑃 (
) = 𝑃(𝑋 < −0.53) = 1 − Φ(0.53)
15
𝑃(𝑋 > 5) = 1 − 𝑃(𝑋 < 5) = 1 − 𝑃 (𝑋 <
5 − 20
) = 1 − 𝑃(𝑋 < −1) = 1 − (1 − Φ(1)) = Φ(1)
15
Adding independent distributions:
𝑋1 ~𝑁(30,100), 𝑋2 ~𝑁(40,150)
𝑌 = 𝑋1 + 𝑋2
𝑌~𝑁(70,250)
𝑋~𝑁(𝜇, 𝜎 2 ) and 𝑌 = 𝑎𝑋 + 𝑏
𝑌~𝑁(𝑎𝜇 + 𝑏, 𝑎2 𝜎 2 )
Subtraction of distributions:
𝑋1 ~𝑁(10,20)
𝑋2 ~𝑁(30,10)
𝑌 = 𝑋1 − 𝑋2
𝑌~𝑁(𝜇1 − 𝜇2 , 𝜎12 + 𝜎22 )~𝑁(−20,30)
The Central Limit Theorem:
𝑋1 , 𝑋2 , … , 𝑋𝑛 are independent and identically distributed with 𝜇 and 𝜎 2 . When n is large (𝑛 ≥
30), then the sum: 𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛 is ~𝑁(𝑛𝜇, 𝑛𝜎 2 )
So even if 𝑋1 , 𝑋2 , … , 𝑋𝑛 aren’t normally distributed, we can use the normal distribution on the
sum.
Sample mean:
1
𝑍 = (𝑋1 + 𝑋2 + ⋯ + 𝑋𝑛 )
𝑛
𝜎2
𝑍~𝑁 (𝜇, 𝑛 )
1
(𝑋 + 𝑋 )
2 1 2 22
𝜎 +𝜎
𝑍~𝑁 (𝜇, 1 22 2 )
𝑍=
Using normal distribution to estimate a discrete distribution:
Continuous can take any value, discrete can take only integers.
𝑃(𝑎 < 𝑥 < 𝑏) ≈ 𝑃(𝑎 − 0.5 < 𝑥 < 𝑏 + 0.5)
If Y follows a normal distribution 𝑌 ∼ 𝑁(𝜇, 𝜎 2 ), then:
𝑏 + 0.5 − 𝜇
𝑎 − 0.5 − 𝜇
𝑃(𝑎 < 𝑥 < 𝑏) = Φ (
)−Φ(
)
𝜎
𝜎
Binomial:
𝑌~𝑁(𝑛𝑝, 𝑛𝑝𝑞)
Poisson:
𝑌~𝑁(𝜆, 𝜆)
Lognormal Distribution:
-parameters 𝜇 and 𝜎 are not the mean and standard deviation.
-if X follows a lognormal with parameters 𝜇 and 𝜎 2 , and 𝑌 = ln 𝑥, then Y follows a normal
distribution with mean 𝜇 and standard deviation 𝜎.
-if X follows a normal distribution with mean 𝜇 and variance 𝜎 2 , and 𝑌 = 𝑒 𝑥 , then Y follows a
lognormal distribution with parameters 𝜇 and 𝜎 2 .
Chapter 8- Multivariate Distribution:
Joint pdf:
𝑓𝑥,𝑦 (𝑥, 𝑦)
Find 𝑝(𝑥 = 𝑥) by adding row 𝑥 = 𝑥 in table.
𝑝(𝑋 = 𝑥) = 𝑓𝑥 (𝑥)
𝑃(𝑌 = 𝑦) = 𝑓𝑦 (𝑦)
𝑓𝑥 (𝑥|𝑌 = 𝑦) =
𝑓𝑥,𝑦 (𝑥, 𝑦)
𝑓𝑦 (𝑦)
𝑓(𝑥, 𝑦) = 𝑓(𝑦|𝑥)𝑓(𝑥)
Independence:
𝑓𝑥,𝑦 (𝑥, 𝑦) = 𝑓𝑥 (𝑥)𝑓𝑦 (𝑦)
= marginal probability functions
𝐸(𝑋𝑌) = 𝐸(𝑋)𝐸(𝑌)
Joint continuous pdf:
𝑓𝑥,𝑦 (𝑥, 𝑦) ≥ 0
Double integral:
Total area must sum to 1
∞
∞
∫ ∫ 𝑓𝑥,𝑦 (𝑥, 𝑦)𝑑𝑥𝑑𝑦 = 1
−∞ −∞
Marginal Continuous probability functions:
∞
𝑓𝑥 (𝑥) = ∫ 𝑓𝑥,𝑦 (𝑥, 𝑦)𝑑𝑦
−∞
∞
𝑓𝑦 (𝑦) = ∫ 𝑓𝑥,𝑦 (𝑥, 𝑦)𝑑𝑥
−∞
∞
𝐸(𝑋) = ∑ 𝑥𝑓𝑥 (𝑥) = ∫ 𝑥𝑓𝑥 (𝑥)𝑑𝑥
−∞
𝐸(𝑋 + 𝑌) = 𝐸(𝑋) + 𝐸(𝑌)
Moment generating function:
𝑀𝑋,𝑌 (𝑠, 𝑡) = 𝐸(𝑒 𝑠𝑥+𝑡𝑦 )
𝐸(𝑋|𝐹) = ∑ 𝑥𝑓𝑥 (𝑥|𝑓)
𝑉𝑎𝑟(𝑌|𝑋) = 𝐸(𝑌 2 |𝑋) − 𝐸(𝑌|𝑋)2
Covariance:
𝐶𝑜𝑣(𝑋, 𝑌) = 𝐸(𝑋, 𝑌) − 𝐸(𝑋)𝐸(𝑌)
𝑉𝑎𝑟(𝑋) = 𝐶𝑜𝑣(𝑋, 𝑋)
𝐶𝑜𝑣(𝑋, 𝑌) = 𝐶𝑜𝑣(𝑌, 𝑋)
𝐶𝑜𝑣(𝑎𝑋 + 𝑏𝑌, 𝑋 + 𝑍) = 𝑎𝐶𝑜𝑣(𝑋, 𝑋) + 𝑎𝐶𝑜𝑣(𝑋, 𝑍) + 𝑏𝐶𝑜𝑣(𝑌, 𝑋) + 𝑏𝐶𝑜𝑣(𝑌, 𝑍)
Independence:
𝐶𝑜𝑣(𝑋, 𝑌) = 0
Correlation Coefficient:
𝐶𝑜𝑣(𝑋, 𝑌)
𝜌=
𝜎𝑥 𝜎𝑦
Chapter 9- Transformations of Random Variables:
Where x is a random variable, 𝑌 = 𝑔(𝑥) is a function of x.
Method of Transformations:
𝑓(𝑦) = 𝑓𝑥 (𝑔−1 (𝑦)) ∗ |[𝑔−1 (𝑦)]′ |
1. 𝑓𝑥 (𝑥) is usually given, as well as 𝑌 = 𝑔(𝑥)
1 −𝑥⁄
𝑓𝑥 (𝑥) =
𝑒 100
100
𝑌 = 1.1𝑋
2. Find 𝑔−1 (𝑦) (aka solve for x in second equation)
𝑌 = 1.1𝑋
𝑋 = 𝑌⁄1.1
3. Use equation
𝑓(𝑦) = 𝑓𝑥 (𝑔−1 (𝑦)) ∗ |[𝑔−1 (𝑦)]′ | = (
1 −(𝑌⁄1.1)⁄
1
1 −𝑌⁄
100 ) ∗ (
𝑒
)=
𝑒 110
100
1.1
110
Method of Distribution Functions:
Use 𝐹𝑦 (𝑦) = 𝑃(𝑌 ≤ 𝑦) and 𝐹𝑥′ (𝑥) = 𝑓𝑥 (𝑥)
1. 𝑓𝑥 (𝑥) and 𝑌 = 𝑔(𝑥) given
𝐹𝑦 (𝑦) = 𝑃(𝑌 ≤ 𝑦)
Substitute Y  𝑃(𝑌 ≤ 𝑦) = 𝑃(1.1𝑋 ≤ 𝑦)
Solve for x 𝑃(𝑋 ≤ 𝑌⁄1.1)
Means: 𝐹𝑥 (𝑌⁄1.1)
2. Use cdf of 𝑓𝑥 (𝑥) = 1 − 𝑒 −(
Notes on SOA 127 Packet:
Average of two variances:
𝑋1 +𝑋2
𝑉𝑎𝑟 (
2
2
) = (12 )𝑉𝑎𝑟(𝑋1 + 𝑋2 )
Average of n variances:
𝑛𝑋
𝑛𝑉𝑎𝑟(𝑋) 𝑉𝑎𝑟(𝑋)
𝑉𝑎𝑟 ( ) =
=
𝑛
𝑛2
𝑛
𝑌⁄ )/100
1.1
= 1 − 𝑒−
𝑌⁄
110
𝑑𝑡
𝑔(𝑦) = 𝑓(𝑡(𝑦)) ∗ | | = 𝑓(𝑡) ∗ 𝑡′
𝑑𝑦
Integration by parts:
𝑏
𝑏
∫ 𝑢 𝑑𝑣 = 𝑢𝑣|𝑏𝑎 − ∫ 𝑣 𝑑𝑢
𝑎
𝑎
Example 1:
𝑏
𝑓(𝑥) = 𝑥𝑒 𝑥 ,
∫ 𝑥𝑒 𝑥 𝑑𝑥 = ∫ 𝑢 𝑑𝑣
𝑎
𝑢=𝑥
𝑑𝑣 = 𝑒 𝑥 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑣 = 𝑒𝑥
implies
𝑏
=
𝑢𝑣|𝑏𝑎
𝑏
− ∫ 𝑣 𝑑𝑢 =
𝑥𝑒 𝑥 |𝑏𝑎
− ∫ 𝑒 𝑥 𝑑𝑥
𝑎
𝑎
Example 2:
𝑓(𝑥) = 𝑥𝑒
−𝑥⁄𝑎
𝑑
,
∫ 𝑥𝑒 −
𝑥⁄
𝑎
𝑑𝑥 = ∫ 𝑢 𝑑𝑣
𝑐
𝑢=𝑥
𝑥
𝑑𝑣 = 𝑒 − ⁄𝑎 𝑑𝑥
implies
𝑑
= 𝑢𝑣|𝑑𝑐 − ∫ 𝑣 𝑑𝑢 = −𝑎𝑥𝑒 −
𝑐
Example 3:
𝑥 𝑥
𝑓(𝑥) = 𝑒 − ⁄𝑎 ,
𝑎
𝑢=𝑥
𝑥
1
𝑑𝑣 = 𝑎 𝑒 − ⁄𝑎 𝑑𝑥
𝑑𝑢 = 𝑑𝑥
𝑥
𝑣 = −𝑎𝑒 − ⁄𝑎
𝑑
∫ 𝑥𝑒 −
𝑥⁄ 𝑏
𝑎|
𝑎
𝑥⁄
𝑎
𝑏
+ ∫ 𝑎𝑒 −
𝑥⁄
𝑎
𝑑𝑥
𝑎
𝑑𝑥 = ∫ 𝑢 𝑑𝑣
𝑐
implies
𝑑
= 𝑢𝑣|𝑑𝑐 − ∫ 𝑣 𝑑𝑢 = −𝑥𝑒 −
𝑐
𝑥⁄ 𝑏
𝑎|
𝑎
𝑑𝑢 = 𝑑𝑥
𝑥
𝑣 = −𝑒 − ⁄𝑎
𝑏
+ ∫ 𝑒−
𝑥⁄
𝑎
𝑑𝑥
𝑎
Deductibles:
An insurance policy has a deductible of d. This means that if x is less than d, the payout is 0, and
if x is more than d, the payout is x-d.
Payout = {
0
𝑥<𝑑
𝑥−𝑑 𝑥 ≥𝑑
∞
𝐸(𝑋) = ∫𝑑 (𝑥 − 𝑑)𝑓(𝑥)𝑑𝑥
∞
𝐸(𝑋 2 ) = ∫𝑑 (𝑥 − 𝑑)2 𝑓(𝑥)𝑑𝑥