Ch3 Operations on one random variable-expectation
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1 Ch3 Operations on one random variable-expectation Previously we define a random variable as a mapping from the sample space to the real line We will now introduce some operations on the random variable. Most of these operations are based on the concept expectation
2 Expectation Expectation is the name given to the process of averaging of a random variable. The followings are equivalents: Expectation or expected value of random variable, which we use the notation E[] The mean value of random variable The statistical average of random variable The following notation are equivalent E[] = Example:3.1-1
3 Expected value of a random variable The everyday averaging procedure used in the above example carries over directly to RV Example: Let be a random variable the has the following sample space values S = {1, 2, 3, 4} Now if the numbers are equally likely to occur or selected = = () 1 ( ) + () 2 ( ) + () 3 ( ) + () 4 4 ( ) = 1 = 2 = 3 = 4 P( = 1) P( = 2) P( = 3) P( =) 4
4 In general then x S = x P( = x ) i i i where each value of the random variable (x i ) is weighted by the Probability P(= x i ) This motivate the concept of expected value or mean value of RV, [ ] E = = x S [ ] E = = i xp(x) i xf (x)dx i if is discrete values if is continuous value with Probability density
5 In the discrete random variable we use the probability mass to weight the random variable. P( = x i ) In the continuous random variable we use the density to weight the random variable. f (x) Observe that f (x)dx represent the probability of the random variable at the interval dx If the random variable is symmetrical about a line x = a f (x + a) = f ( x + a) E [ ] = a Example 3.1-2
6 Expected value of a Function of a random variable Assume a random variable which has the following values and probabilities = {1, 2, 3} P(=1) = P(=2) = P(=3) = Now define the random variable Y = Y = {1, 4, 9} 1 2 P(Y=1) = P(Y=4) = P(Y=9) = E[Y] = () 1 ( ) + () 4 ( ) + ( 9) ( ) Y= 1 Y= 4 Y= 9 P( Y= 1) P( Y = 4) P( Y = 9) P( = 1) P( = 2) P( = 3) yi {1,4,9} = y P(Y= y ) y i {1,4,9} = yp(= x ) i i i i
7 In general then [ ] Eg() = N g(x)p(x) i i i=1 for discrete random variable [ ] E g(x) = g(x)f (x)dx for continuous random variable
8 Conditional Expectation We define the conditional density function for a given event B = { b} f (x b) = f (x) x < b b f (x)dx 0 x b we now define the conditional expectation in similar manner [ ] b E B = xf (x B)dx = xf (x B)dx + xf b (x B)dx f (x) constant = F (b) x < b b = x dx + x 0 dx b b f (x)dx 0 = b x b xf (x)dx b f (x)dx constant = F ( b)
9 Moments The expected value defined previously as [ ] E = = xf (x)dx if is continuous value with probability density f (x) Is called the 1st moment of the random variable with probability density f (x) The word moment is used because a similar form exist in static were the 1st moment represent the center of gravity
10 Example Assume a mass is distributed on one dimension x as shown below were m(x) is the mass density function is Then we can calculate the 1st moment or center of gravity M as M = b a b x m(x) dx m(x) dx a total mass In the probability case the total area under the density function is unity or 1
11 The expected value of the function of random variable was defined as [ ] E g(x) = g(x)f (x)dx Let the function g() defined as n g() = n = 0, 1, 2,... Then we can define the n th moment m n as n n m n = E = xf(x)dx
12 Clearly m = E = x f (x)dx = 1 The area of the function f (x) E1 [] =1 f (x)dx = m = E = xf (x)dx = The expected value of
13 Central Moments Another type of moments of interest is the central moment defined as n n µ n = E ( ) = (x ) f (x)dx the m n moments are expected values about the origin however the central moment m n is moment or expected value about the mean or average The area of the function µ = E ( ) = ( ) f (x)dx = 1 f (x) E1 [] =1 f (x)dx = 1 1 µ 1 = E ( ) = E[] E[] = = 0 were we have used the fact that E[ a ] = a constant
14 Variance and skew The second central moment µ 2 is so important, that it is given the name variance and have the special notation 2 σ x Thus the variance is given by x 2 σ = µ = E ( ) = (x ) f (x)dx σ ( the positive square root of the variance) is called the standered deviation of RV σ is a measure of the spread of the random variable about its mean or average
15 σ is a measure of the spread of the random variable about its mean or average f(x) 1 f(x) 2 σ 1 σ 2 σ > σ 1 2 The spread of f(x) is more than the spread of 2 1 f(x) ( σ > σ ) 1 2
16 variance can be found from a knowledge of first and second moments as follows [ ] σ = µ = E 2 + = E 2E x 2 = E = m m Example
17 Properties of the variance σ Let c be a constant and be a RV 2 (1) σ c The probability density function of a constant deterministic number is a delta function with spread zero f (x) δ (x c)
18 (2) σ = σ 2 2 x + c x The variance does not change by shifting the random variable, the spread will remain the same, on the other hand shifting effect the mean only (3) σ = c σ cx x 3 The third central moment µ is a measure of the 3 = E ( ) a symmetry of f (x) about x = = m 1 It will be called the skew of the density function If f (x) symmetric about x =, it has zero skew (i.e µ 3 = 0 ) µ n = 0 for all odd n. 3 The normalized third central moment µ 3 σ is known as the x skewness or coefficient of skewness of the probability density function
19 Useful Inequalities A useful tool in some probability problems are some inequalities such as Chebychev s inequality and Markov s Inequality. Chebychev s Inequality It state that for a RV, { } 2 2 P є σ є for any ε > 0
20 Proof { } 2 2 P є σ є ε + ε P є = P є and є { } { } { } = P + є and є є = f (x)dx + f (x)dx = f (x)dx +є x є but since x σ = (x ) f (x)dx (x ) f (x)dx 2 2 (x ) ε 2 2 x { } f (x)dx = P { } 2 2 P є σ є
21 Another form of Chebychev s Inequality is { } 2 2 P < 1 (σ ) { } 2 2 P є σ є A consequence of that if 2 σ 0 for a random variable, then P{ < } 1 or P{ = } 1 In other words, if the variance of the RV approach zero, the probability approaches 1, that will equal its mean.
22 Markov s Inequality Let be a non negative random number, then P{ a } E[ ] a a > 0
23 3.4 Transformations of A Random Variable Let be a random variable with a known distribution F (x) and a known density f (x). Let T(.) be a transformation or a mapping or a function that maps the R.V into Y as Y = T() The problem is to find F Y (y) and f Y (y). You can view the problem as a block box problem
24 In general can be a discrete, continuous or mixed random variable and the Transformation T can be Linear Non-linear Segmented Staircase We will consider three cases: (1) is continuous and T is continuous and monotonically increasing or decreasing
25 (2) is continuous and T is continuous nonmonotonic (3) is discrete and T is continuous
26 Monotonic Increasing Transformations of a Continuous Random Variable A transformation T is called monotonically increasing if T(x 1 ) < T(x 2 ) for any x 1 < x 2. Assume that T is continuous and differentiable at all values of x for which f (x) 0. y = T( x ) or x = T ( y )
27 The events {Y y 0 } and { x 0 } are equivalent. P {Y y 0 } = P { x 0 } Differentiating both sides with respect to y 0 and using Leibniz rule Let = 0 y x T ( y ) Y = { } { } F ( y ) = P Y y = P x = F ( x ) Y f ( y) dy f ( x) dx β ( u) Gu ( ) = H( xudx, ) α ( u) β ( u) dg( u) dβ( u) dα( u) H ( x, u) = H[ β( u), u] H[ α( u), u] + dx du du du u α ( u)
28 Now the LHS, d dy y dy f ( y) dy = f ( y ) f ( y )(0) Y Y 0 Y 0 0 dy 0 y dfy ( y) dy dy 0 = f Y ( y ) 0 The RHS d dy o 1 1 0= 0 0= 0 x T [ y ] x T [ y ] dx f ( x) dx= f [ x ] f ( x )(0) dy0 df ( x) dx dy 0 = dt f ( T ( y )) ( y ) dy0 dt f ( y ) = f ( T ( y )) Y ( y ) dy0
29 Since the results apply for any y 0, fy( y) = f T ( y) dx or fy( y) = f ( x) dy 1 1 dt ( y) dy
30 Monotonic Decreasing Transformations of a Continuous Random Variable A transformation T is called monotonically decreasing if T(x 1 ) > T (x 2 ) for any x 1 < x 2. { } { } F ( y ) = P Y y = P x = 1 F ( x ) Y y x T ( y ) 1 0 0= 0 f ( y) dy = 1 f ( x) dx Y d dy y0 1 1 dt y 0 fy( y) dy = fy( y) = 0 f[ T ( y0)] 0 dy 0 ( )
31 1 dt ( y0) since is negative (monotone decreasing) dy0 1 1 ( ) ( ) ( ) dt fy y f y T y f( x) dx = = dy dy Then we conclude that dx f ( y) = f ( x) Y dy for both increasing and decreasing monotonic transformation.
32 Nonmonotonic Transformations of a Continuous Random Variable A transformation may not be monotonic(increasing and decreasing) in the more general case it can be both increasing and decreasing as shown below The event { Y y 0 } may correspond to more than one event of the random variable.
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