• Distribution function– this is for the discrete case, the distribution function is thus denoted by F(x) and is expressed as

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

1. P (a < X < b) = F (b) – F (a)
2. P (a ~ X <b) = P[X = a] + [F (b) – F (a)]
• P (a < X ~ b) = [F (b) – F (a)] – P[X = b]
1. P (a < X < b) = F (b) – F (a) + P [X = a] – P [X = b]

Continuous case

1. P (a < X < b) = P (a < X < b) = P (a < X ~ b) = P (a < X < b) = F (b) – F (a)
2. F (-∞) = Lt x à-∞ F(x) = 0

F (∞) = Lt  xàF(x) = 1

• F(x) ->F(y) whenever x < y.
1. F (a) – F (a – 0) = P[x = a] and F (a + 0) = F (a)
2. For continuous case, F'(x) = j (x) ~ 0 => F(x) is non-decreasing function.

• Mean/expectation- consider X as a random variable. Then the mean/ expectation is defined as

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:

1. E[X + Y] = E[X] + E[Y]

1. E [cX] = c E[X]

• E[c] = c and E [X + c] = E[X] + c

1. If X and Y are independent then E (XY) = E(X). EM

1. Physically, expectation signifies the center of mass of the probability distribution.

• Variance- it is the probability distribution expressed as below:

σ2= V[X] = Σx (x- µ) 2 p(x)

=ʃ-∞ (x – µ)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:

1. V[aX + b] = a2 V[X]

1. Physically the variance signifies the moment of inertia of the probability mass distribution in a line, perpendicular to the line of the distribution.

Example 1. For the following distribution

 X 1 2 3 4 5 p(x) 0.1 k 0.2 3k 0.3

1. Find out the value of k
2. Calculate the mean and variance
• Find out the distribution function

Solution:

1. Since this is a pmj we have,

Σ p(x) = 1

= > 0.1 + k + 0.2 + 3k + 0.3

= > 4k + 0.6 = 1

= > 4k = 0.4

= > k = 0.1

1. µ = mean = Σ x. p (x)

= 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.

• Distribution function is given below as follows:
 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.

1. A r.v. X will probably have no moments, in spite of mgf existing.

Example:

f(x) = 1/(x+1)(x+2),   x=0,1,2…… (The reader can verify)

1. A r.v. X can also have moments, in spite of its mgf is unsuccessful to produce the moments.

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 and Kurtosis:

Skewness =1= 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 = µ42,                     ɣ2 = ᵦ2 – 3

For, ᵦ2 = 3 this type is called the platykurtic

2> 3 this type is known as mesokurtic

2< 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 + 6µ212 – 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]

Σx=0 etx qx = p Σx=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

• Mean Deviation (M.D.)

M.D. = ʃ-∞ [x- µ] f(x)dx                                      (in case of continuous)

= Σx [x – µ]. P {X=x}                                          (in case of discrete)

• Median- The median is defined as the point that splits the whole distribution into two equal parts. In case ofpdf, the total portion is being parted into two equal halves by the median. Therefore after creation the solution we get the following:

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

1. A random variable X consists of the distribution as given below

(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).

1. The probability distribution as given below:

 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).

1. A random variable X has the probability distribution:
 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.

1. Assume the probability distribution given below:

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

1. The Maxwell-Boltzmann distribution is given below:

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.

1. The range of a random variable namely X contains of the points I, 2, …, up to nprovided information is P[X = i] is proportional to 1/ i(i+1). Now evaluate the distribution function of x. Calculate P[3 < X S n] and P[X > 5].

1. Three balls are taken out randomly without replacement from acontainerwhich consists of 4 red and 6 white balls. Now let’s consider X as a random variable which indicates the total number of red balls taken out from the container, then draw a table that represents the probability distribution of X. calculate the expectation also.

1. Calculate the estimated value and variance of the number of times the heads appears when two coins are tossed.

1. Let the probability density function be

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.

1. Calculate the mgf of the following:

(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).

1. In a delivery of 7 items there were three defective items. Then random 4 items were selected. Then find out the expected number of defective items?

1. The diameter of an electric cable wasexpected to be anuninterrupted random variable X with pdf

f(x) = 6x(1 – x), 0 <x < 1.

Calculatebsuch that P[X <b] = P[X >b].

1. (a)   0.2

(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

1. (a) 2,

(b) 2/3, 1/18

1. Expected value = 1, · Variance= 1/2
2. (I) 0.33, (ii) 0.414, (iii) 0, (iv) 0.732, (v) 0.3172, (vi) 2.4,

(vii) F (x) = ¼ (x+1)2, -1≤x≤1

=0, elsewhere

11.12/7

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