F(x) = P[X ≤ x ]
The distribution for the above function will be all follows:
x | -1 | 0 | 1 |
F(x) | 0.2 | 0.6 | 1 |
For continuous case, the function distribution van be defined like,
F(x) = P[X ≤ x] = ʃ f(x) dx
Hence for the above example,
F(x) = x ʃ-∞f(x) dx = 0ʃ-∞f(x) dx + x ʃ0 f(x) dx
= 0 + x ʃ0 1. dx = x
Properties:
Discrete case
Continuous case
F (∞) = Lt xà∞ F(x) = 1
µ = E [X] = Σ x.p (x) (discrete case)
= ∞ʃ –∞x f(x) dx (continuous case)
It has got an exception which is known as the ‘Population mean’.
Properties:
σ2= V[X] = Σx (x- µ) 2 p(x)
=∞ʃ-∞ (x – µ)2 f(x) dx
σ2 = E [(X – µ)2] = E[X2] – {E[X]}2
= ΣX x2 . p(x) – µ2
= P (M1/A)
P(M1).P(A/M1)
= ———————————————————————–
P(M1).P(A/M1)+P(M2).P(A/M2)+P(M3).P(A/M3)
0.2 X 0.3
= ———————————————————————
0.2 X 0.03 + 0.45 X 0.05 + 0.34 X 0.04
= 0.14
Properties:
Example 1. For the following distribution
X | 1 | 2 | 3 | 4 | 5 |
p(x) | 0.1 | k | 0.2 | 3k | 0.3 |
Solution:
Σ p(x) = 1
= > 0.1 + k + 0.2 + 3k + 0.3
= > 4k + 0.6 = 1
= > 4k = 0.4
= > k = 0.1
= 1(0.1) + 2(0.1) + 3(0.2) + 4(0.3) + 5(0.3) = 3.6
σ2 = Variance = Σ (x – µ)2 . p(x)
= (1 – 3.6) (0.1) + (2 – 3.6)2 (0.1) + (3 – 3.6)2 (0.2)+ (4 – 3.6f (0.3) + (5 – 3.6)2 (0.3)
=1.64.
X | 1 | 2 | 3 | 4 | 5 |
P(x) | 0.1 | 0.2 | 0.4 | 0.7 | 1 |
Example 2:
Find out the mean, variance and distribution of the function pdf
f(x) = ax2, 0 .::;, x .::;, 1
= 0, elsewhere.
Solution: since this is a pdf, then
∞ʃ –∞x f(x) dx = 1
=1ʃ0x f(x) dx = 1
= a[ x3/ 3]10 = 1
a = 3
So the pdf can be taken as
f(x) 3×2, 0 < x > 1
= 0, elsewhere
µ = mean = 1ʃ0 x f(x) dx = 31ʃ0 x3dx = 3 [x4/4]10 = ¾
σ2 = Variance =1ʃ0 (x- ¾)2. 3x2.dx
= 3/16 1ʃ0 [16x4 – 24x3 + 9x2] dx
= 3/16 [16/5 – 24/4 + 9/3] = 3//80
Distribution function
= 3 xʃ0 x2 dx,0 ≤ x ≤ 1
=[x3 , 0 ≤ x ≤ 1
[0 , elsewhere.
Moments and moment generating function:therth instant of the mean is defined as
µr = E[(X – µ)r] = Σx (x – µ)r. P[ X = x]
=∞ʃ-∞(x- µ)r f(x) dx
Take a look at the discrete case now,
Mx (t) = Σ pi et(xi– t) , where p(x) is expressed by P.
= Σ pi [1 + t(xi – a ) + t2/2! (xi-a)2+….]
= Σ pi + t Σ pi (xi – a) + t2/2!Σ pi (xi-a)2
= 1 + t µ’1 + (t2/2!) µ’2 + …………+ (tr/r!) µ’1+…..
So the µ’ = coefficient of tr/r! in the expression mx(t).
On the other hand we have,
µr’ = [d’/dt’. Mx(t)]t=0
Thus the relation of raw moments and central moments are like:
µ2 = µ’2 -µ2 = variance
µ4 = µ’4 – 4µ’3.µ + 6µ’2 µ2 – 3µ4
(Thus the moment that is acquired from a distribution (discrete/ continuous) is known as “population moments”)
Note. I. in case in the mgf if the point an absent, then it can be taken as zero.
Example:
f(x) = 1/(x+1)(x+2), x=0,1,2…… (The reader can verify)
Example 3. The pdf of Rayleigh scattering is expressed as:
µ3 = µ‘3 – 3 µ‘2. µ + 2 µ3
f(x) = [x/a2 exp (-x2– 2a2) x≥0
, x<0
Find out the distribution function, variance and mean.
Solution:
For x ≥ 0, 1ʃ0x/a2 .exp (-x2/2a2)dx = 1ʃ0exp (-x2/2a2).d (x2/2a2)
= 1 – exp (-x2/2a2)
Distribution function – [1-0 exp (-x2/2a2), x ≥ 0
X < 0
µ = 1ʃ0(x2/2a2).exp (-x2/2a2)dx
Putting, (x2/2a2)=t , so xdx= a2dt
= a√2 ∞ʃ0e-1 t1/2 dt
= a√2 ∞ʃ0e-1 t3/2-1 dx
= a√2 √3/2
= a√2 ½
Now, = µ = 1ʃ0 f(x)dx (considering the point a=0)
Variance = µ’2 – µ2
= 2a2a2/2
= (2- a2
Skewness =ᵦ1= µ32/ µ23
ɣ1 = √ᵦ1
For the symmetric scattering, ᵦ1 = 0. If ᵦ1 > 0 then the dispersion is known as positive skew.
ᵦ1 < 0 , in this case the distribution is known as a negative skew.
Kurtosis = ᵦ2 = µ4/µ2, ɣ2 = ᵦ2 – 3
For, ᵦ2 = 3 this type is called the platykurtic
ᵦ2> 3 this type is known as mesokurtic
ᵦ2< 3 this type is known as the leptokurtic
Example 4: a continuous random variable
f(x) = 3×2, 0 < x< 1
Beginning the calculation with the first four central moments,β1 and β2
µ’1 = 1ʃ0x.f(x) dx = 3 1ʃ0x3 dx = 3 / 4
µ’2 = 1ʃ0x2.f(x) dx = 3 1ʃ0x4 dx = 3 /5
µ’3 = 1ʃ0x3 f (x) dx = 3 1ʃ0x5 dx = 1 / 2
µ’4 = 1ʃ0x4.f(x) dx = 3 1ʃ0x6 dx = 3 / 7
µ’1 = µ1 = 3 / 4 = 0.75
µ2 = µ’2 – ( µ1)2 = 3/5 – (3/4)2 = 3 / 80 = 0.0375
µ3 = µ’3 – 3µ’2. µ1 + 2µ13
= 1 /2 – 3. 3 /5. ¾ +2 = – 1/160 = 0.00625
µ4 = µ’4– 4µ’3. µ1 + 6µ2’.µ12 – 3µ14
3/ 7 – 4.1 /2.3+ 6.3/5.9/16 – 3.81/256
= 0.00435
So β1 = µ32 / µ23 = (0.00625)2 / (0.00375)3 = 3.09
So now, therefore β1> 0, where the distribution is a skewed positive.
β2 = µ4 / µ22 = 0.00435/(0.0375)2 = 3.09
Since β2 > 3, the distribution is leptokurtic.
Example 5. Find out the moment producing function of the distribution given below.
P[X = x1= q2p, x = 0, 1, 2, ….., 0 <p .:S 1, q = .1 – p
= 0, otherwise.
So calculate the the mean and variance.
Solution. The given arrangement is evidently a discrete distribution.
Mx (t) = E [ etx]
Σ∞x=0 etx qx = p Σ∞x=0(et.q)x
P (1-qe. t1)-1
p/1-qet, this is mgf
µ1’ = [d/dt M(t)]t=0
= [d/dt. p(1-qet)-1]t=0 = [p q et (1-q et)-2] t=0
= pq (1-q)-2 = pq/p2 = p/q (therefore, p=1-q)
µ2’ = [d/dt M(t)]t=0
pq [d/dt {e’(1-qet)-2}]t=0 (taken from the previous part)
= pq [ et ( 1- q et ) -2-2et(1-qet)-3. (-q et)
= pq[ (1 – q)-2+ 2q(1 – qr)3] = .p/q + 2q2/p2
Mean = µ1’= q/p
Variance = µ2 = µ2’- (µ1’)2 =q/p + 2q2/p2 – q2/p2 = q/p + q2/p2 = pq + q2/q2 = q/p3
M.D. = ∞ʃ-∞ [x- µ] f(x)dx (in case of continuous)
= Σx [x – µ]. P {X=x} (in case of discrete)
Mʃ-∞ f(x)dx= ½ or ∞ʃM f(x)dx= ½
(h) Mode. Mode can be defined as the value of x, where f(x)carries the maximum value and is expressed likef'(x) = 0, f “(x) < 0. This lies in the interlude of the arrangement.
(i) Quartiles. In pdf, the quartiles Q1 and Q3 are expressed as following
Q2ʃ-∞ f(x)dx= 1/4 and ∞ʃQ3 f(x)dx= 1/4
Note: now let’s take a look on a couple of special probability
(i) There remains achance that a specific magnitude (K) might exceed up to, P[X > K] = p0 and this isknown as the ‘Exceedance Probability’.
(ii) Median Exceedance Probability can be well explained as the value that exceeds 50 in a example of evaluations of the exceedance probability of certain specific magnitude.
Example 6.Find the median for the arrangement shown below.
f(x) = 3x2, 0 <x <1.
Solution. Let us assume M as the median, then
f(x) dx = 1/2
3 x2dx = 1/2
= 1/2
M2 =1/2 = M = (1/2)1/3 = 0.079
Example 7. Find outthe mean deviation of the distribution given below.
X | 0 | 1 | 2 | 3 | 4 |
P(X) | 0.1 | 0.3 | 0.4 | 0.5 | 0.6 |
Solution. Here in this problem we se, µ = Σp(x)
= 0 + 1(0.3) + 2(0.4) + 3(0.1) + 4(0.1)
= 0.3 + 0.8 + 0.3 + 0.4 = 1.8
Hence, M.D = Σ| x-µ | p(x)
= 0 – 1.8 . 0 + l – 1.8 (0.3) + 2 – 1.8 (0.4) + 3 – 1.8 (0.1) + 4 – 1.8 (0.1) = 0.66.
Example 8. Find out the quartiles of the following:
f(x) = 3×2, 0 <x < 5. 1.
Solution. For lower quartile, Q2ʃ0 f(x)dx= 1/4
Q2ʃ0 x2dx= 1/4
[ x3]Q10 = 1/4
Q13 ¼ = q1 =(1/4)1/3
For upper quartile, 3. 1ʃQ3 x2 dx= 1/4
[ x3]1Q3 = 1/4
1-Q33 = 1/4 = Q33 = 3/4 = Q3 (3/4)1/3
PROBLEMS
(a) Calculate the value of k.
(b) Find out the mean and variance.
(c) Evaluate the mean deviation.
(d) solve: P(l < x < S 3), P(x > 1).
x | 1 | 2 | 3 | 4 | 5 | 6 |
P(x) | a | 2a | 3a | 4a | 5a | 6a |
(i) Find the value of a.
(ii) Calculate the distribution function.
(iii} evaluate the mean and variance.
(iv) Solve P(X<4), P(2 < X < 5).
x | -2 | -1 | 0 | 1 | 2 |
p(x) | 0.2 | 0.1 | 0.1 | 0.3 | 0.3 |
Find out the first four central moments.
f(x) = kx, O<x<l
(a) Calculate the value of k.
(b) Bring out the mean and variance.
(c) Solve to find the median, Q1 and Q3
f(x) = 4a. x2 e-ax2 , 0< x < ∞ ,a > 0
In the above expression, a = m/2kT, m = Mass, T = Temperature (K), k = Boltzmann constant and x = Speed of a gas molecule.
(a) Now prove that this is a pdf.
(b) Determine the mean and variance.
f(x) = 0.5 (x + 1), -1 < x <1
= 0, elsewhere
Now evaluate the (i) Mean, (ii) Median, (iii) Q1, (iv)Q3, (v) β1, (vi) β2 (vii) Distribution function.
(i) F(x) = 1/b-a, a <x <b
= 0, elsewhere.
(ii) p(x) = e– λ λx /x , x = 0,1, 2….
Hence find the mean and variance in (ii).
f(x) = 6x(1 – x), 0 <x < 1.
Calculatebsuch that P[X <b] = P[X >b].
ANSWERS
(b) 1.1, 1.69
(c) 1.1
(d) 0.7, 0.5.
2 (i) a = 0.1
(iii) 3.9, 2.89
(iv)0.4, 0.2
3 µ1, = 0, µ2 = 2.24, µ3 = -1.75, µ4 = 9.03
(b) 2/3, 1/18
(vii) F (x) = ¼ (x+1)2, -1≤x≤1
=0, elsewhere
11.12/7
12.b=1/2
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