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STAT3600 Chapter 4 Maghsoodloo
CONTINUOUS RANDOM VARIABLES
Let X be a continuous rv with range space R (^) x (a subset of real numbers). Since the Pr that the rv, X, takes on a single specific value in R (^) x is zero, then Prs over a continuum must be defined only over intervals within R (^) x. The function, f(x), is said to be a Pr density function (pdf) over R (^) x iff (1) f(x) > 0 for all x in R (^) x, (2) (^) ∫ R x
f (x)dx = 1 , (3) f(x) must be
(piecewise) continuous, (4) f(x) = 0 for all x ∉ R (^) x, and (5) P(a < X < b) = P(a < X < b) = P(a < X < b) = P(a < X < b) =
∫ ab f^ (x)dx.^ Note that all these last 4 Prs are equal because the
pr for a single point (or any set of measure 0) for any continuous rv is identically zero.
Example. The repair time of certain engines, X in hours, of an automobile is given by the pdf f(x) = a + bx , 1 ≤ x ≤ 4 hours and f(x) = 0 elsewhere. The interval [1, 4 hrs] is called the range space of X. (a) Determine the constants a and b such that f(x) is a Pr density function (pdf). The graph of the above function is given below. Because the area under a triangle is equal to (1/2)base×height, then to
make the total area equal to 1 we require that (1/2)3×h = 1 → h = 2/3.
x
Note that the modal point of the above pdf is MO = 1 and f(MO) = h. To obtain the Pdf itself, we insert the points (1, 2/3) and (4, 0) into f(x) = a + bx. 2/3 = a + b and 0 = a + 4b → − 2/3 = 3b → b = − 2/9, and a = +8/9 → f(x) = = ( − 2/9)x + 8/9, Rx = [1, 4 hours].
∫ R x
f (x)dx =
4 1
2x (^8) dx
∫ =^
2 4 ⎡⎣ (^) − x + 8x ⎤⎦ (^) 1 =
9 [16^ −^ (^ −^ 1+ 8)]
= (16 − 7)/9 = 1 → Thus, f(x) is a pdf.
(b) Compute the Pr that the next repair time lies in the interval [1.5, 2 hours].
1 hour 4 hrs
f(x)
(1, h=f(MO))
FX (t)
t
P(X > t) = the shaded area
hours.
P( X > 3 hours ⎪X ≥ 2) = P(XP(X^^ >> 3)2) =^11^ −−^ P(XP(X^ <≤ 3)2) = X X
1 F (3)
1 F (2)
− =^
THE EXPECTED VALUE OF A CONTINUOUS RV
Definition: Recall that if X is a discrete rv over the range Rx with pmf p(x), then the population mean of the rv, X, is defined as its expected value and is given by μ = E(X) = ∑ x p(x), where the sum extends over Rx.
If X is continuous with pdf f(x), then
μ = E(X) = R x
∫ x f (x)dx.
Since an integral is simply summing over a continuum and f(x)dx represents the pr element at x (actually over an interval of length dx), then the form of E(X) for a continuous rv is identical to that of the discrete case.
Example (continued). Compute the average repair time per job (or mean time to complete a maintenance task).
μ = E(X) =
4 1
∫ x( 2x/9 + 8/9)dx- =
3 2 4 1
2x 4x 27 9
= 2.00.
THE VARIANCE OF A CONTINUOUS RV
As before, the population variance is the long-run (or weighted) average of deviations from the population mean ( μ ) squared, i.e.,
V(X) = σ^2 = E[(X − μ ) 2 ] = (^) ∫ −μ R x
( x )^2 f(x)dx = x
2 2 R ∫ (x^ −^^2 μ^ x^ + μ )f (x)dx
R x
x 2 f(x)dx − μ^2. E(X^2 ) =^4 1 ∫ x ( 2x/9 + 8/9)dx- =
4 3 4 1
x 8x 18 27
= 4.50 → σ^2 x = V(X) = 0.50 → CV (^) x = 0.50 2 = 35.35534%
THE MOMENTS OF A RANDOM VARIABLE
Definition: Let X be a rv with the range space R (^) x and let c be any known constant. Then the kth^ moment of X about the constant c is defined as Mk (X) = E[ (X − c)k^ ]. (12)
In the field of statistics only 2 values of c are of interest: c = 0 and c = μ. Moments about c = 0 are called origin moments and are denoted by μk/, i.e., μk/^ = E(Xk^ ), where c = 0 has been inserted into equation (12). Moments about the population mean, μ, are called central moments and are denoted by^ μk, i.e,
then
μ 3 = E[(X − μ) 3 ] = (^) ∫ − μ + μ −μ R x
(x^3 3 x^23 x^23 )f(x)dx
= μ′ 3^ − 3 μ′ 2^ μ + 2μ^3 (14)
For our example, we have shown that μ = μ/ 1 = 2, μ′ 2^ = 4..
μ′ 3^ = E(X^3 ) =
(^4 ) 1 ∫ x ( 2x/9 + 8/9)dx- =
5 4 4 1
2x 2x 45 9
⎢ −^ + ⎥
→ μ 3 = μ′ 3^ − 3 μ′ 2^ μ + 2 μ^3 = 11.20 − 3(4.5)(2) +2(2^3 ) = 0.
If X is measured in hours, the units of μ 3 are expressed in terms of hours 3. To obtain a unit-less measure of asymmetry (for comparative purposes), we standardize μ 3 to obtain the coefficient of skewness (most authors refer to this coefficient simply as skewness) given below: α 3 = μ 3 / σ^3. α 3 = 0.20/(0.5) 1.5^ = 0.565685425 > 0 which is unit-less and shows that our pdf is positively skewed (long tail on the RHS), and thus μ > x (^) 0.50 > MO It is not common for the value of α 3 = μ 3 /σ^3 to lie outside the interval [−2, 2].
(4) The 4th central moment is a measure of Kurtosis (peaked- ness in the middle and heavy Prs at the tails) and is given by (for a continuous rv) μ 4 = (^) ∫ −μ R x
( x )^4 f(x)dx = x
4 3 2 2 3 4 R ∫ (x^ −^ 4x^ μ +^ 6x^ μ^ −^ 4x^ μ^ + μ )f (x)dx
This last expression after simplifying reduces to
μ 4 = μ′ 4^ − 4 μ′ 3^ μ + 6 μ′ 2^ μ^2 − 3 μ^4 (15)
For our example, μ′ 4^ =
(^4 ) 1
∫ x ( 2x/9 + 8/9)dx- =
6 5 4 1
x 8x 27 45
= 30.
→ μ 4 = μ′ 4^ − 4 μ′ 3^ μ + 6 μ′ 2^ μ^2 − 3 μ^4 = 30.20 − 4(11.2)(2)+6(4.5)(2^2 )
− 3(2^4 ) = 0.60.
Again in order to obtain a unit-less measure of kurtosis, we 1st^ standardize μ 4 to obtain a (unitless ) measure of kurtosis defined as α 4 = β 2 = μ 4 /σ^4. we then normalize the value of α 4 = β 2 = μ 4 /σ^4 by 3 and use the terminology kurtosis = (μ 4 / σ^4 ) − 3 = (0.6)/0.5^2 − 3 = 2.40 − 3 = − 0.60 = β 4. All Triangular distributions in the universe have a Kurtosis β 4 = α 4 −3 = μ 4 / σ^4 − 3 = − 0.60000 (Platykurtic).
The percentiles of Continuous rv X (i.e., inverting the
cdf)
The median is a point on the abscissa such that FX (x0.5) = 0.50. For our example,
of F(x) is given by
F −^1^ (x) = 4 − 3 1 − x , 0 ≤ x ≤ 1
because F[ F −^1^ (x) ] =^19 [−(4 − 3 1 − x ) 2 +8(4 − 3 1 − x ) − 7 ] =
x. Similarly, F −^1^ [F(x)] = 4 − 3 1 − F(x) =
4 − 3 1 − ( x + 8x 7)/9- 2 - = 4 − 9 − ( - x + 8x^2 - 7) =
4 − x 2 - 8x +16 = 4 − (4 - x)^2 = x.
Examples of Inverse Functions
(a) f(x) = x 2 , x ≥ 0 → f −^1 (x) = x
(b) f(x) = x1/3^ → f −^1 (x) = x^3 , − ∞ < x < ∞
(c) f(x) = e x^ → f −^1 (x) = ln(x)
(d) f(x) = sin(x) → f −^1 (x) = arcsin(x)
(e) f(x) =^12 ln( 11^ −+ xx) → f −^1 (x) = tanh(x) =
ex e x
ex e x
−^ −
(f) f(x) = 10 x^ → f −^1 (x) = log 10 (x)
The Gamma Function
Γ(n) = n^1 x 0
x e dx
∞ (^) − −
∫ =^ 0 u dv
∞
∫ =^ u v^ ] 0 ∞^ −^ 0 v du
∞
u = xn^ −^1 , dv = e −^ x^ dx One integration by parts yields
Γ(n) = 0
⎡ (^) xn −^1^^ ( − e − x) ⎤^ ∞
⎣ ⎦ + (n^ −^ 1)^ ∫
∞ (^) − − − 0
x( n^1 )^1 e xdx =(n − 1) Γ(n −1)
or Γ(n + 1) = n Γ(n) (21)
Thus, if we integrate (20) a total of (n − 1) times by parts, we obtain Γ(n) = (n − 1) (n − 2)(n − 3)…1.Γ(1)
But Γ(1) =^0 x 0
x e dx
∞ (^) −
0
⎡ (^) − e^ − x ⎤^ ∞ ⎣ ⎦ = 1, and hence
Γ(n) = (n − 1) (n − 2)(n − 3)…1 =(n − 1)! →Γ(1) =(1− 1)! = 0! = For example, Γ(5) = 4! = 24, while Γ(7) = 6! = 720, etc. Further, it can be proven that Γ(1/2) = π. If n = 3/2, from equation (21), Γ(n + 1) = n Γ(n) so that Γ(3/2)
= Γ( (^) 21 + 1) = (^) 21 Γ( (^) 21 ) = (^) 21 π. Similarly, Γ(5/2) = Γ( (^) 23 + 1) = (^) 23
Γ(3/2) = 3 π / 4. You should verify for yourself that Γ(7/2) =
15 π / 8, etc. In the expression for the gamma function Γ(n), let x = λt → dx = λ dt. Substituting the transformation x = λt into equation (20) yields
Γ(n) = ∫
∞ (^) − −λ λ λ 0
( t)n^1 e t dt → 1 = ∞∫^ λ − −λ
Γ
λ 0
( t)n^1 e tdt (n ) ,^ (23)
f(x) = 0.0004 e−0.0004 x^ is called the exponential density function at a rate of λ = 0.0004 failures (or Poisson events) per hour. (b) Compute the Pr that the computer will survive a 10-hour flight.
P(X > 10 hours ) = 0.0004x 10
0.0004e dx
∞ (^) −
∫ = e−0.0004^ ×^10 = e−0.004^ =
0.996007989344 ≅ roughly 4 failures in 1000 flights. The above Pr is called the reliability (or survival Pr) of the PC at 10 hours. This implies that the failure Pr of the PC during the 10-hour flight is P(X ≤ 10) = 1 − 0.996007989344 = 0.003992010656, while the reliability at 10 hours is equal to R(at 10 hours) =
(c) Obtain the cdf (cumulative distribution function) of TTF, X, at time t.
FX (t) = P(X ≤ t) = (^) ∫ λ e −^ λ dx =
t 0
x
t
0
x 1
e ⎥
⎥ ⎦
⎤ ⎢
⎢ ⎣
⎡ −
−λ
= 1 − e −λ t^ = 1 − e−0.0004t
Note that once the cdf of a rv is obtained, then all Prs can be computed using the cdf. For example, we may use the cdf to recompute the reliability of the PC in part (b) above. R(10 hours) = P(TTF > 10 ) = 1 − P(X ≤ 10) = 1 − FX (10) = 1 − (1 − e−^ λt^ ) = e−^ λt^ = e−^10 λ^ = e−0.004^ = 0.996008 =
Pr[X(10 hours) = 0] (d) Use the cdf to compute the Pr that the PC will fail within the interval (10, 20 hours). P(10 ≤ X ≤ 20 hours) = P(X ≤ 20) − P(X ≤ 10) = F(20) − F(10) = = (1 − e−^20 λ^ ) − (1 − e−^10 λ^ ) = e−^10 λ^ − e−^20 λ^ = = R(10) − R(20) = e−^ 0.004^ − e−^ 0.008^ = 0.0039761. (e) Suppose it is known that on a certain flight the PC has already lasted 7 hours. What is the Pr that its lifetime will go beyond 10 hours?
P(X > 10⏐ X > 7) = P(X^ P(X^ >^^10 ∩> 7)X^ > 7) = P (XP(X^^ >> 10 ) 7) =
10 7
e
e
− λ − λ = e
− 3 λ
= P(X > 3 hours ⏐X > 0) = P(X > 3 hours) = R(3) = e−^ 0.0012^ =
The above developments show that the exponential density function is memory-less because the conditional pr that the PC’s lifetime exceeds 10 hours given that it has lasted 7 hours is the same as the unconditional pr that the PC lasts 3 hours from time zero. Thus, in general the exponential pdf has the following memory-less property
P(X > a+b ⏐ X > a) = P(X > b) = e −λ b^. The discrete analogue of the exponential density is the Geometric distribution g(x; p). To my knowledge, these are the only two memory-less statistical distributions.
Thus,
R(t) = P(T > t) = P[X(t) = 0] = e −λ t^ = 1 − P(T < t) = 1 − F(t) →
F(t) = 1 − e −λ^ t → f(t) = dF(t)/ dt = λ e −λ t
The above developments show that if the number of occurrences of an event is Poisson distributed, then the intervening time, T, of the next Poisson event measured from the last occurrence is exponentially distributed.
- n > 1 Example (continued) Consider a 2-PC standby system that guides an aircraft each PC with a failure rate of λ = 0.0004/hour. Compute the system reliability for a mission of 10-hour flight. R(10) = P(Time to the 2 nd^ failure measured form zero > 10 hours) = P(T 2 > 10 hours) = P[X(10 hours) < 1 failure] =
=
1 x 0. x 0
(0.004) (^) e x!
−
∑ = 0.99999202^ ≅^ 8 failures in a
million. Note that X(10 hours) denotes the number of failures occurring during 10 hours, which has a Poisson pmf with mean μ = λt = 0.0004×10 = 0.004 failures per 10 hours. The system MTTF = E(T 2 ) = n/λ = 2/0.0004 = 5000 hours, and
V(T 2 )= σ^2 = n/λ^2 = 2/(0.0004^2 ) = 12500000 hours^2 → σ = 3535.53391→ CV(T 2 ) = 70.711%. Now consider a 3-PC (or 3-unit) standby system that guides an aircraft each PC with a failure rate of λ = 0.0004/hour. Compute the system reliability for a mission of 10-hour flight. R(10) = P(T 3 > 10 hours) = P[X(10 hours) < 2 failures] =
=
(^2) − x 0
x e 0. 004 x!
( 0. 004 ) = 0.9 (^7) 8936528 = 10.635 failures in a billion.
The system MTTF = E(T 3 ) = n/λ = 3/0.0004 = 7500 hours, and V(T 3 )= σ^2 = n/λ^2 = 18.75 x 10^6 hours^2 → σ = 4330.127 → CV(T 3 ) = 57.735%. We can also compute the above reliability at 10 hours directly from the gamma pdf as follows:
R(10) = P(T 3 > 10 hours) =^3 1 x 10 (3)(^ x)^ e^ dx
∞ (^) λ (^) − − λ
∫ Γ λ ; Let y =^ λx
→ dy = λ(dx) → R(10) =^2 y 0.
(^1) (y) e dy 2!
∞ (^) −
∫ =^^12 {^ −^ y e2 -y^ 0.
y 0.
2(y)e dy
∞ (^) −
∫ } =^^12 { (0.004)^2 e^ −^ 0.004^ + [^ 2ye^ y 0.
− −^ ⎤∞
y 0.
e dy
∞ (^) −
∫ ] } =^^12 { (0.004)^2 e^ −^ 0.004^ +^ 2(0.004)e − 0.004^ +^ 2e −^ 0.004 } =
e −^ 0.004^ + (0.004)e − 0.004^ +^1 2 (0.004)^
(^2) e − 0.004 (^) = 0.9 (^78936528)
These last 2 moments show that as n → ∞, the distribution of Tn approaches normality because the 3 rd^ and 4th^ standardized moments of a normal distribution are α 3 = 0 and α 4 = E(Z^4 ) = 3. Further, because λ determines the standard deviation, it is called the scale parameter while n is called the shape parameter because n determines the skewness and kurtosis. Exercise 33 p.69 (a) λ = 0.20 failures/Week FT (2) = P(Time to the 1st^ downtime < 2 weeks) = (T < 2 weeks) (^2) 0.20t 0
∫ 0.20e^ − dt =^
0.20t^2
e 0
− −^ ⎤
⎦ = 1^ −^
e − 0.20(2)^ = 1 − e − 0.40^ = 0.32968.
Let X(2 weeks) = No. of failures during 2 weeks. FT (2) = P(T < 2) = P[X(2 weeks) ≥ 1] = 1 − P[X(2) = 0] =1 − Poisson pmf(at x = 0 and λt = 0.40)= 1 − p(0; λt = 0.40) = 1
− e − 0.40^ = 0.32968.
(b) P[X(6) = 4] = ( t)^ x x!
λ e −λ t = (0.20 6)^4
× e − 1.20 =0.
P[X(6) ≤ 6] =
6 x 1. x 0
(1.20) e
x!
−
(c) P(T < 4) = FT (4) =
(^4) 0.20t 0
∫ 0.20e^ − dt = 1^ −^ e − 0.80^ = 0.550671 =
P[X(4 weeks) ≥ 1] = 1 − P[X(4) = 0] = 1 − p(0; λt = 0.80 ) (e) Compute the Pr that the time to the 4 th^ failure (measured from the last failure, i.e., T 4 ) exceeds 9 weeks.
P(T 4 > 9 weeks) = P[X(9 weeks) ≤ 3] = F(3; λt = 1.80) =
Compute the Pr that the time to the 6 th^ failure is less than 15 weeks. P(T 6 ≤ 15 weeks) = P[X(15 weeks) ≥ 6] = 1 − P[X(15 weeks) ≤ 5]
= 1 − F(5; λt = 3) =1 −
5 x 3 x 0
3 e
x!
−
∑ = 1^ −^ 0.916082058 = 0.083918.
