Jan 16, 2016
0.
0.1
-50 -40 -30 -20 -10 0 10 20 30 40 50
X = 2*Bin(300,1/2) – 300E[X] = 0
0.
0.1
-50 -40 -30 -20 -10 0 10 20 30 40 50
Y = 2*Bin(30,1/2) – 30E[Y] = 0
0.
0.1
-50 -40 -30 -20 -10 -38 10 20 30 38 48
Z = 4*Bin(10,1/4) – 10E[Z] = 0
0.
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.
-50 -40 -30 -20 -10 0 10 20 30 40 50
W = 0E[W] = 0
• Is there a good parameter that allow to distinguish between these distributions?
• Is there a way to measure the spread?
• The variance of X, denoted by Var(X) is the mean squared deviation of X from its expected value = E(X):
Var(X) = E[(X-)2]. The standard deviation of X, denoted
by SD(X) is the square root of the variance of X.
SD(X) = Var(X).
2 2Var(X) = E(X ) E(X) .Proof:
E[ (X-)2] = E[X2 – 2 X + 2]
E[ (X-)2] = E[X2] – 2 E[X] + 2
E[ (X-)2] = E[X2] – 22+ 2
E[ (X-)2] = E[X2] – E[X]2
Claim: Var(X) = E[(X-)2].
1. Claim: Var(X) >= 0.
Pf: Var(X) = (x-)2 P(X=x) >= 0
2. Claim: Var(X) = 0 iff P[X=] = 1.
Theorem: X is random variable on sample space S, and P(X=r) it’sprobability distribution. Then for any positive real number r:
(proof in book)
2
( )(| ( ) | )
V XP X E X r
r
In words: the probability of finding a value of X farther away from the meanthan r is smaller than the variance divided by r^2.
r
What is the probability that with 100 Bernoulli trials we find more than89 or less than 11 successes when the prob. of success is ½.
X counts number of successes.EX=100 x ½ =50
V(X) = 100 x ½ x ½ = 25.
P(|X-50|>=40)<=25/40^2 = 1/64
“k standard deviations”
Let be random variables that for
, ,
What is the probability of getting 25 or fewer heads in 100 coin tosses?