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    Eng Lecture 31 - Expected Value

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    Eng Lecture 31 - Expected Value

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    © All Rights Reserved
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    16 views9 pages

    Eng Lecture 31 - Expected Value

    Uploaded by

    QWERTYUIOP

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    You are on page 1of 9

    Probability Theory and Statistics 114

    Prof. Willie Brink


    Applied Mathematics, Stellenbosch University

    Lecture 31
    Expected value of a continuous random variable

    1/9
    Previously...

    A continuous random variable X is defined by a probability density function f (x),


    with the following properties :

    • f (x) ≥ 0 for all x ∈ (−∞, ∞)

    Z ∞
    • f (x) dx = 1
    −∞
    Z b
    • P{a ≤ X ≤ b} = f (x) dx
    a

    2/9
    Example 1c

    Suppose X is the lifetime (in hours) of a certain radio tube, with density
    (
    0, x ≤ 100
    f (x) = 100
    x 2 , x > 100

    Probability that exactly 2 out of 5 such tubes will have to be replaced within 150 hours?

    Let Y be the number of tubes out of 5 that have to be replaced within 150 hours.
    Then Y is a binomial random variable, with n = 5 and p = P{X ≤ 150}.

    Z 150 Z 150
    p= f (x) dx = 100x −2 dx = 1
    3
    −∞ 100

    5 1 2 2 3
      
    P{Y = 2} = 2 3 3 = 0.3292

    3/9
    Cumulative distribution function

    Consider a random variable X with probability density function f (x).


    The cumulative distribution function associated with X is
    Z a
    F (a) = P{X ≤ a} = f (x) dx.
    −∞

    Note that P{a ≤ X ≤ b} = F (b) − F (a). y = f (x)

    d
    It also follows from the above that F (a) = f (a).
    da
    a b x

    4/9
    Example 1d

    Let X be a continuous random variable, with density function fX and distribution function FX .
    Find the density function of the random variable Y = 2X .

    FY (a) = P{Y ≤ a}
    = P{2X ≤ a}
    = P{X ≤ 2a }
    = FX ( 2a )

    Differentiate both sides with respect to a:

    fY (a) = 12 fX ( 2a )

    5/9
    5.2 Expected value and variance

    The expected value of a continuous random variable X with density function f (x) is
    defined as follows :
    Z ∞
    E [X ] = x f (x) dx.
    −∞

    Note: it is the continuous analogy to the expected value of a discrete random variable,
    which we defined as
    X
    E [X ] = x P{X = x}.
    x

    6/9
    Example 2a

    Find E [X ] if the density function of X is given by


    (
    2x, 0 ≤ x ≤ 1
    f (x) =
    0, otherwise

    Z ∞
    E [X ] = x f (x) dx
    −∞

    Z 1
    = 2x 2 dx
    0

    2
    =
    3

    7/9
    Proposition 2.1

    If the probability density function of X is f (x), then for any real function g (·),
    Z ∞
    E [g (X )] = g (x) f (x) dx.
    −∞

    Note: the analogy to the discrete case is again quite clear:


    X
    E [g (X )] = g (x) P{X = x}.
    x

    8/9
    Example 2b

    The probability density function of X is given by


    (
    1, 0 ≤ x ≤ 1
    f (x) =
    0, otherwise

    Find E [e X ].

    Z ∞
    E [e X ] = e x f (x) dx
    −∞

    Z 1
    = e x dx
    0

    = 1.7183

    9/9

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