**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 = ^{0}ʃ_{-∞}f(x) dx + ^{x }ʃ_{0 }f(x) dx

= 0 + ^{x }ʃ_{0 }1. dx = x

**Properties:**

Discrete case

- P (a < X < b) = F (b) – F (a)
- P (a ~ X <b) = P[X = a] + [F (b) – F (a)]

- P (a < X ~ b) = [F (b) – F (a)] – P[X = b]

- P (a < X < b) = F (b) – F (a) + P [X = a] – P [X = b]

Continuous case

- P (a < X < b) = P (a < X < b) = P (a < X ~ b) = P (a < X < b) = F (b) – F (a)
- F (-
_{∞) }= Lt_{ x }_{à}_{-∞ }F(x) = 0

F (∞) = Lt_{ x}_{à}_{∞ }F(x) = 1

- F(x) ->F(y) whenever x < y.

- F (a) – F (a – 0) = P[x = a] and F (a + 0) = F (a)
- 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:**

- E[X + Y] = E[X] + E[Y]

- E [cX] = c E[X]

- E[c] = c and E [X + c] = E[X] + c

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

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

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

σ^{2}= V[X] = Σ_{x} (x- µ)^{ 2 }p(x)

_{ =}^{∞}ʃ_{-∞ (x }– µ)^{2 }f(x) dx

σ^{2 }= E [(X – µ)^{2}] = E[X^{2}] – {E[X]}^{2}

= Σ_{X }x^{2 }. p(x) – µ^{2}

= P (M_{1}/A)

P(M_{1}).P(A/M_{1})

= ———————————————————————–

P(M_{1}).P(A/M_{1})+P(M_{2}).P(A/M_{2})+P(M_{3}).P(A/M_{3})

0.2 X 0.3

= ———————————————————————

0.2 X 0.03 + 0.45 X 0.05 + 0.34 X 0.04

= 0.14

**Properties:**

- V[aX + b] = a2 V[X]

- 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 |

*Find out the value of k**Calculate the mean and variance*

*Find out the distribution function*

**Solution:**

- 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

- 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

=^{1}ʃ_{0}x f(x) dx = 1

= a[ x^{3}/ 3]^{1}_{0 }= 1

a = 3

So the *pdf* can be taken as

f(x) 3×2, 0 < x > 1

= 0, elsewhere

µ = mean = ^{1}ʃ_{0 }x f(x) dx = 3^{1}ʃ_{0 }x^{3}dx = 3 [x^{4}/4]^{1}_{0 }= ¾

σ2 = Variance =^{1}ʃ_{0 }(x- ¾)^{2}. 3x^{2}.dx

= 3/16 ^{1}ʃ_{0} [16x^{4 }– 24x^{3 }+ 9x^{2}] dx

= 3/16 [16/5 – 24/4 + 9/3] = 3//80

Distribution function

= 3 ^{x}ʃ_{0 }x^{2 }dx,0 ≤ x ≤ 1

=[x^{3} , 0 ≤ x ≤ 1

[0 , elsewhere.

**Moments and moment generating function:**therth instant of the mean is defined as

µ_{r }= E[(X – µ)^{r}] = Σ_{x} (x – µ)^{r}. P[ X = x]

=^{∞}ʃ_{-∞}(x- µ)^{r} f(x) dx

Take a look at the discrete case now,

M_{x} (t) = Σ p_{i} e^{t(x}_{i}^{– t)} , where p(x) is expressed by P.

= Σ p_{i }[1 + t(x_{i} – a ) + t^{2}/2! (x_{i}-a)^{2}+….]

= Σ p_{i }+ t Σ p_{i }(x_{i} – a) +_{ t}^{2}/2!Σ p_{i }(x_{i}-a)^{2}

= 1 + t µ’_{1 }+ (t^{2}/2!) µ’_{2} + …………+ (t^{r}/r!) µ’_{1}+…..

So the µ’ = coefficient of t^{r}/r! in the expression mx(t).

On the other hand we have,

µ_{r}’ = [d’/dt’. M_{x}(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 }µ^{2} – 3µ^{4}

(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.

- 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)

- 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/a^{2 }exp (-x^{2}– 2a^{2}) x≥0

, x<0

Find out the distribution function, variance and mean.

**Solution:**

For x ≥ 0, ^{1}ʃ_{0}x/a^{2} .exp (-x^{2}/2a^{2})dx = ^{1}ʃ_{0}exp (-x^{2}/2a^{2}).d (x^{2}/2a^{2})

= 1 – exp (-x^{2}/2a^{2})

Distribution function – [^{1-}_{0} exp (-x^{2}/2a^{2}), x ≥ 0

X < 0

µ = ^{1}ʃ_{0}(x^{2}/2a^{2}).exp (-x^{2}/2a^{2})dx

Putting, (x^{2}/2a^{2})=t , so xdx= a^{2}dt

= a√2 ^{∞}ʃ_{0}e^{-1} t^{1/2} dt

= a√2 ^{∞}ʃ_{0}e^{-1} t^{3/2-1} dx

= a√2 √3/2

= a√2 ½

Now, = µ = ^{1}ʃ_{0} f(x)dx (considering the point a=0)

Variance = µ’_{2} – µ^{2}

= 2a^{2}a^{2}/2

= (2- a^{2}

**Skewness and Kurtosis:**

Skewness** =**ᵦ_{1}= µ_{3}^{2}/ µ_{2}^{3}

ɣ_{1} = √ᵦ_{1}

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 = ᵦ_{2 }= µ_{4}/µ_{2}, ɣ_{2} = ᵦ_{2} – 3

For, ᵦ_{2 }= 3 this type is called the platykurtic

ᵦ_{2}> 3 this type is known as mesokurtic

ᵦ_{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,β_{1}* and *β_{2}

µ’_{1 }= ^{1}ʃ_{0}x.f(x) dx = 3 ^{1}ʃ_{0}x^{3} dx = 3 / 4

µ’_{2 }= ^{1}ʃ_{0}x^{2}.f(x) dx = 3 ^{1}ʃ_{0}x^{4 }dx = 3 /5

µ’_{3 }= ^{1}ʃ_{0}x^{3} f (x) dx = 3 ^{1}ʃ_{0}x^{5} dx = 1 / 2

µ’_{4 }= ^{1}ʃ_{0}x^{4}.f(x) dx = 3 ^{1}ʃ_{0}x^{6} dx = 3 / 7

µ’_{1 }= µ_{1} = 3 / 4 = 0.75

µ_{2 }= µ_{’2 }– ( µ_{1})^{2} = 3/5 – (3/4)^{2} = 3 / 80 = 0.0375

µ_{3} = µ’_{3 }– 3µ’_{2}. µ_{1} + 2µ_{1}^{3}

= 1 /2 – 3. 3 /5. ¾ +2 = – 1/160 = 0.00625

µ_{4} = µ’_{4}– 4µ’_{3}. µ_{1} + 6µ_{2}^{’}.µ_{1}^{2} – 3µ_{1}^{4}

3/ 7 – 4.1 /2.3+ 6.3/5.9/16 – 3.81/256

= 0.00435

So β1 = µ_{3}^{2} / µ_{2}^{3} = (0.00625)^{2} / (0.00375)^{3 }= 3.09

So now, therefore β_{1}> 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 = x^{1}= q^{2}p, 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 [ e^{tx}]

Σ^{∞}_{x=0} e^{tx} q^{x} = p Σ^{∞}_{x=0}(e^{t}.q)^{x}

P (1-q^{e}. t^{1})^{-1}

p/1-q^{et}, this is mgf

µ_{1}’ = [d/dt M(t)]_{t=0}

= [d/dt. p(1-qe^{t})^{-1}]_{t=0} = [p q e^{t }(1-q e^{t})^{-2}] t=0

= pq (1-q)^{-2} = pq/p^{2} = p/q (therefore, p=1-q)

µ_{2}’ = [d/dt M(t)]_{t=0}

pq [d/dt {e’(1-qe^{t})^{-2}}]_{t=0 }(taken from the previous part)

= pq [ e^{t} ( 1- q e^{t} ) ^{-2}-2e^{t}(1-qe^{t})^{-3}. (-q e^{t})

= pq[ (1 – q)^{-2}+ 2q(1 – qr)^{3}] = .p/q + 2q^{2}/p^{2}

Mean = µ_{1}’= q/p

Variance = µ_{2 }= µ_{2}’- (µ_{1}’)^{2} =q/p + 2q^{2}/p^{2 }– q^{2}/p^{2 } = q/p + q^{2}/p^{2 } = pq + q^{2}/q^{2} = q/p^{3}

* *

**Mean Deviation (M.D.)**

M.D. = ^{∞}ʃ_{-∞} [x- µ] f(x)dx (in case of continuous)

= Σ_{x} [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 of*pdf,*the total portion is being parted into two equal halves by the median. Therefore after creation the solution we get the following:

^{M}ʃ_{-∞} f(x)dx= ½ or ^{∞}ʃ_{M} 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

^{Q}_{2}ʃ_{-∞} f(x)dx= 1/4 and ^{∞}ʃ_{Q3} 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) = 3x^{2}, 0 __<__x __<__1.

* ***Solution. **Let us assume M as the median, then

f(x) dx = 1/2

3 x^{2}dx = 1/2

= 1/2

M^{2} =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, **^{Q2}ʃ_{0} f(x)dx= 1/4

^{Q2}ʃ_{0} x^{2}dx= 1/4

[ x^{3}]^{Q1}_{0} = 1/4

Q_{1}^{3} ¼ = q_{1} =(1/4)^{1/3}

**For upper quartile**, 3. ^{1}ʃ_{Q3 }x^{2 }dx= 1/4

[ x^{3}]^{1}_{Q3} = 1/4

1-Q^{3}_{3} = 1/4 = Q^{3}_{3} = 3/4 = Q_{3} (3/4)^{1/3 }

** **

**PROBLEMS**

- 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).

- 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).

- 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.

- 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

- 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.

- 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].

- 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.

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

- 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) Q_{1}, (iv)Q_{3,} (v) β_{1, } (vi) β^{2} (vii) Distribution function.

- Calculate the mgf of the following:

(i) F(x) = 1/b-a, a __<__x __<__b

= 0, elsewhere.

(ii) p(x) = e^{– λ} λ^{x} /x , x = 0,1, 2….

Hence find the mean and variance in (ii).

- 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?

- 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].

** **

** **

**ANSWERS**

- (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 µ^{1}, = 0, µ2 = 2.24, µ3 = -1.75, µ4 = 9.03

- (a) 2,

(b) 2/3, 1/18

- Expected value = 1, · Variance= 1/2
- (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

12.b=1/2

** **