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P(X = x) = 1. |
all x |
Notes:
For example, if X is the rv 'the score obtained when one die is tossed', then X can takes on the values 1, 2, 3, 4, 5, 6, and therefore x = 1, 2, 3, 4, 5 or 6.
'X = 2' is read as 'the score obtained is 2', whereas 'x = 2' means 'x is 2'.
'P(X = 2)' is read as 'the probability that the score obtained is 2'.
A function which is responsible for allocating probabilities of a random variable X is called the probability density function (pdf) of X.
Pdf may be given as a table or expressed as a function of x.
F(x) = | P(X £ x) | ||||
=
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So for a given real number x, F(x) is the sum of the probabilities up to and including the events assign to x.
Properties of F(x)
The mean value or expectation of a drv X, written as m or E(X), is given by
E(X) = |
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Note: If the pdf of X is symmetrical about the central value c, then E(X) = c.
In general, if g(X) is any function of the random variable X, then
E[g(X)] = |
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Properties of E
The variance of a rv X, denoted by Var(X), is defined as
The standard deviation of X, denoted by s, is the square root of Var(X):
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s = Ö |
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So Var(X) = s2.
Computational formula for Var(X):
Properties of Var
If X and Y are any two random variables,
then for any constants a and b, If X and Y are also independent, then |
Note: Confusions often arises over the following:
2X
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- twice the value of one observation of X |
X1 + X2
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- the sum of two independent obsevations of X |
E(2X) =
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2E(X) |
E(X1 + X2) =
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E(X1) + E(X2) = 2E(X) |
Var(2X) =
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22Var(X) = 4Var(X) |
Var(X1 + X2) =
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Var(X1) + Var(X2) = 2Var (X). |
A die, with numbers 1 to 6, is weighted such that the probability of obtaining a score is proportional to the score. Let W denotes the score obtained when the die is tossed once. Find the pdf, cdf, expectation and variance of W. If X denotes three times the score when the die is tossed once, and Y denotes the sum of the scores when the die is tossed thrice, what are the expectation and variance of X and Y.
Solution:
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= 1 | ||||
k(1 + 2 + ... + 6)
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= 1 | ||||
k
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= 1/21 |
The pdf of W is P(W = w) = w/21 for w = 1, 2, ..., 6.
For w = 1, 2, ..., 6,
F(w) = |
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=
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(1 + ... + w)/21 | ||||
=
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w(w + 1)/42 |
The cdf of W is F(x) = w(w + 1)/42 for w = 1, 2, ..., 6.
E(W) = |
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=
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(12 + 22 + ... + 62)/21 | ||||
=
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13/3 |
E(W2) = |
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=
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(13 + 23 + ... + 63)/21 | ||||
=
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21 |
Var(W) = | E(W2) - E2(W) |
=
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21 - (13/3)2 |
=
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20/9 |
X = 3W | Y = W1 + W2 + W3 |
E(X) = 3E(W) = 13 | E(Y) = E(W1) + E(W2) + E(W3) = 13 |
Var(X) = 32Var(W) = 20 | Var(Y) = Var(W1) + Var(W2) + Var(W3) = 20/3 |