Continuous Random Variables
Suppose that the range space of X is made up of a very large finite number
of values, say all values x in the interval 0 S x S I of the form 0, 0.01,
0.02, ... , 0.98, 0.99, 1.00. With each of these values is associated a nonnegative
number p(xi) = P(X = xi), i = l , 2, . . . , whose sum equals 1. We have pointed out before that it might be mathematically easier to idealize
the above probabilistic description of X by supposing that X can assume all
possible values, 0 ≤ x ≤ I. If we do this,
what happens to the point probabilities p(xi)? Since the possible values of X are non countable,
we cannot really speak of the i value of X, and hence p(xi) becomes meaningless.
What we shall do is to replace the function p,
defined only for x1. x2, • • • , by a function f defined (in the present context) for all values of x, 0≤x≤1. The properties of Eq. (4.3) will be replaced by f(x)≥0 and
\[\int_{0}^{1}{f\left( x \right)}dx=1\].
Let us proceed formally as follows.
\[\int_{0}^{1}{f\left( x \right)}dx=1\].
Let us proceed formally as follows.
Definition. X is said to be a continuous random variable if there exists a function
f, called the probability density function (pdf) of X, satisfying the following
conditions:
\[\begin{align}
& \left( a \right)f\left( x \right)\ge 0 \\
& \left( b \right)\int_{-\infty }^{+\infty }{f\left( x \right)dx=1} \\
& we have P\left( a\le X\le b \right)=\int_{a}^{b}{f\left( x \right)dx} \\
\end{align}\]
assume all values in some interval (c, d) where c and d may be - ∞ and + ∞, respectively.
The stipulated existence of a pdf is a mathematical device which has considerable intuitive
appeal and makes our computations simpler. In this connection it should again be
pointed out that when we suppose that X is a continuous random variable, we are dealing
with the idealized description of X.
(b) P(c < X < d) of the pdf f between x = c and x =d.
(c) It is a consequence of the above probabilistic description of X that for any specified
value of X, say xo, we have P(X = xo) = 0, since P(X = xo) = Σ(x) dx = 0.
This result may seem quite contrary to our intuition. We must realize, however, that if
we allow X to assume all values in some interval, then probability zero is not equivalent
with impossibility. Hence in the continuous case, P(A) = 0 does not imply A = Ï•,
the empty set. (See Theorem 1.1.) Saying this more informally, corisider choosing a
point at random on the line segment {x|0 ≤ x ≤ 2}. Although we might be willing to
agree (for mathematical purposes) that every conceivable point on the segment could be
the outcome of our experiment, we would be extremely surprised if in fact we chose
precisely the midpoint of the segment, or any other specified point, for that matter.
When we state this in precise mathematical language we say the event has "probability
zero." In view of these remarks, the following probabilities are all the same if X is a
continuous random variable:
P(c ≤ X ≤ d), P(c ≤ X < d), P(c < X ≤ d), and P(c < X < d).
(d) Although we shall not verify the details here, it may be shown that the above
assignment of probabilities to events in Rx satisfies the basic axioms of probability
(Eq. 1.3), where we may take {x / - ∞ < x < + ∞ } as our sample space.
(e) If a function f* satisfies the conditions,.{f*(x)≤ 0, for all x, and
\[\int\limits_{-\infty }^{+\infty }{{{f}^{*}}}(x)dx=k\],
where K is a positive real number (not necessarily equal to 1), then f* does not satisfy all
the conditions for being a pdf. However, we may easily define a new function, say f, in
terms of f* as follows:
f(x)=f*(x)/k for all x
Hence f satisfies all the conditions for a pdf.
(f) If X assumes values only in some finite interval [a, b], we may simply set f(x) = 0
for all x∉ [a, b]. Hence the pdf is defined for all real values of x, and we may require
that
\[\int\limits_{-\infty }^{+\infty }{{{f}^{*}}}(x)dx=k\],
Whenever the pdf is specified only for certain values of x, we shall suppose that it is zero elsewhere.
(g) f (x) does not represent the probability of anything! We have noted before that
P(X = 2) = 0, for example, and hence/(2) certainly does not represent this probability.
Only when the function is integrated between two limits does it yield a probability. We
can, however, give an interpretation of f(x) t::..x as follows. From the mean-value theorem
of the calculus it follows that
\[P\left( x\le X\le x+\vartriangle x \right)=\int_{x}^{s+\vartriangle s}{f\left( s \right)ds}=\vartriangle xf\left( \xi \right),where(x\le \xi \le x+\vartriangle x)\]
If △x is small, f(x) △x equals approximately P(x ≤ X ≤ x + △x). (If f is continuous
from the right, this approximation becomes more accurate as △x = 0.)
(h) We should again point out that the probability distribution (in this case the pd0
is induced on Rx by th.. underlying probability associated with events in S. Thus, when
we write P(c < X < d), we mean, as always, P[c < X(s) < d], which in turn equals
P[s I c < X(s) < d], since these events are equivalent. The above definition, Eq. (4.8),
essentially stipulates the existence of a pdf f defined over Rx such that
\[p[s|c
We shall again suppress the functional nature of X and hence we shall be concerned only
with Rx and the pdf f.
(i) In the continuous case we can again consider the following analogy to mechanics:
Suppose that we have a total mass of one unit, continuously distributed over the interval
a ≤ x ≤ b. Then f(x) represents the mass density at the point x and ∑f(x) dx represents
Y = H(X) is a continuous random variable, and it will be our task to obtain
its pdf, say g.
The general procedure will be as follows:
(a) Obtain G, the cdf of Y, where G(y) = P( Y ≤ y), by finding the event A
(in the range space of X) which is equivalent to the event { Y≤ y}.
(b) Differentiate G(y) with respect to y in order to obtain g(y).
(c) Determine those values of y in the range space of Y for which g(y) > 0.
Continuous Random Variables
The most important (and most frequently encountered) case arises when X
is a continuous random variable with pdf f and His a continuous function. HenceY = H(X) is a continuous random variable, and it will be our task to obtain
its pdf, say g.
The general procedure will be as follows:
(a) Obtain G, the cdf of Y, where G(y) = P( Y ≤ y), by finding the event A
(in the range space of X) which is equivalent to the event { Y≤ y}.
(b) Differentiate G(y) with respect to y in order to obtain g(y).
(c) Determine those values of y in the range space of Y for which g(y) > 0.
the total mass contained in the interval c ≤ x ≤ d.
EXAMPLE 1
Suppose that x has pdf
f(x) = 2x, 0 < x < 1,
= 0, elsewhere.
Let H(x) = 3x + I. Hence to find the pdf of
Y = H(X) we have
G(y) = P( Y ≤ y) = P(3 X + 1 ≤ y)
= P(X ≤ (y - 1)/3)
\[{{\int_{0}^{\frac{\left( y-1 \right)}{3}}{2xdx=\left[ \frac{\left( y-1 \right)}{3} \right]}}^{2}}\]
Thus
g(y) = G'(y) = 2/9(Y - 1).
Since f(x) > 0 for 0 < x < l, we find that
g(y) > 0 for 1 < y < 4.
Note: The event A, referred to above, equivalent to the event { Y ≤ y} is simply
{X ≤ (y )1/3}.
There is another, slightly different way of obtaining the same result which will
be useful later on. Consider again
G(y) = P( Y ≤ y) = P ( X ≤ y -1/3) = F ({y-1}/3)
where F is the cdf of X; that is,
F(x) = P(X ≤ x).
In order to evaluate the derivative of G, G'(y), we use the chain rule for differentiation
as follows:
dG(y)/dy = dG(y)/du *du/dy , where u= (y-1)/3
Hence
G'(y) = F'(u) · 1/3 = f(u) ·1/3 = 2 ({y-1}/3). 1/3
as before. The pdf of Y has the graph
Theorem 1.
Let X be a continuous random variable with pdf f, where f(x) > 0 for a < x < b. Suppose that y = H(x) is a strictly monotone (increasing or decreasing) function of x. Assume that this function is differentiable (and hence continuous) for all x. Then the random variable Y defined as Y = H(X) has a pdf g given by,
'
g(y) = f(x) |dx/dy|
where x is expressed in terms of y. If His increasing, then g is nonzero for
those values of y satisfying H(a) < y < H(b). If H is decreasing, then g
is nonzero for those values of y satisfying H(b) < y < H(a).
Proof: (a) Assume that His a strictly increasing function. Hence
G(y) = P( Y ≤ y) = P(H(X) ≤ y)
= P(X ≤ H-1(y)) = F(H-1(y)).
Differentiating G(y) with respect to y, we obtain, using the chain rule for derivatives,
Thus
dG(y)/dy = dG(y)/dx . dx/dy where x = H-1(y).
G'(y) = dF(x)/dx dx/dy = f(x)dx/dy
(b) Assume that H is a decreasing function. Therefore
G(y) = P(Y ≤ y) = P(H(X) ≤ y) = P(X ≤ H-1(y))
= 1 - P(X ≤ H-1(y)) = 1 - F(H-1(y)).
Proceeding as above, we may write
dG(y)/dy = dG(y)/dx .dx/dy = d[1-F(x)]/dx . dx/dy = -f(x)dx/dy
Note: The algebraic sign obtained in (b) is correct since, if y is a decreasing function of x, x is a decreasing function of y and hence dx/dy < 0. Thus, by using the absolute value sign around dx/dy we may combine the result of (a) and (b) and obtain the final form of the theorem.
EXAMPLE 2
Let us reconsider Examples 5.6 and 5.7, by applying Theorem 1
(a) For Example we had f(x) = 2x, 0 < x < 1, and y = 3x + I. Hence x = (y - 1)/3 and dx/dy = 1/3 Thus g(y) = 2[(y - 1 )/3}1/3 = 4/9(y - 1 ),
1 < y < 4, which checks with the result obtained previously.
(b) In Example 1 we had f(x) = 2x, 0 < x < l , and y = e-x. Hence
x = -in y and dx/dy = - 1 /y. Thus g(y) = -2 (in y)/y, 1/e< y<1 p="" which="">again checks with the above result.
If y = H(x) is not a monotone function of x we cannot apply the above
method directly. Instead, we shall return to the general method outlined above.
The following example illustrates this procedure.
EXAMPLE 3
Suppose that
f(x) = 1/2 -1 < x < 1,
= 0, elsewhere.
Let H(x) = x 2. This is obviously not a monotone function over the interval [ -l, l ] Hence we obtain the pdf of Y = x 2 as follows:
<1 p="" which=""> G(y) = P( Y ≤ y) = P(x 2 ≤ y)
= P( -√y ≤ x≤ √y)
= F(yy) - F(-yy),
where F is the cdf of the random variable X. Therefore
g(y) = G'(y) = f(√y)/2√y - f(-√y)/(-2√y)
<1 p="" which=""> =1[f(√y)' + f(- √y)]/2√y'
<1 p="" which=""> Thus g(y) = (l/2√y) (1/2 + 1/2) = l/2√y, 0 < y < l. (
The method used in the above example yields the following general result.
Theorem 5.2. Let X be a continuous random variable with pdf f Let Y = x 2•
Then the random variable Y has pdf given b
<1 p="" which="">g(y)=<1 p="" which="">1[f(√y)' + f(- √y)]/2√y'
EXAMPLE 1
Suppose that x has pdf
f(x) = 2x, 0 < x < 1,
= 0, elsewhere.
Let H(x) = 3x + I. Hence to find the pdf of
Y = H(X) we have
G(y) = P( Y ≤ y) = P(3 X + 1 ≤ y)
= P(X ≤ (y - 1)/3)
\[{{\int_{0}^{\frac{\left( y-1 \right)}{3}}{2xdx=\left[ \frac{\left( y-1 \right)}{3} \right]}}^{2}}\]
Thus
g(y) = G'(y) = 2/9(Y - 1).
Since f(x) > 0 for 0 < x < l, we find that
g(y) > 0 for 1 < y < 4.
Note: The event A, referred to above, equivalent to the event { Y ≤ y} is simply
{X ≤ (y )1/3}.
There is another, slightly different way of obtaining the same result which will
be useful later on. Consider again
G(y) = P( Y ≤ y) = P ( X ≤ y -1/3) = F ({y-1}/3)
where F is the cdf of X; that is,
F(x) = P(X ≤ x).
In order to evaluate the derivative of G, G'(y), we use the chain rule for differentiation
as follows:
dG(y)/dy = dG(y)/du *du/dy , where u= (y-1)/3
Hence
G'(y) = F'(u) · 1/3 = f(u) ·1/3 = 2 ({y-1}/3). 1/3
as before. The pdf of Y has the graph
Theorem 1.
Let X be a continuous random variable with pdf f, where f(x) > 0 for a < x < b. Suppose that y = H(x) is a strictly monotone (increasing or decreasing) function of x. Assume that this function is differentiable (and hence continuous) for all x. Then the random variable Y defined as Y = H(X) has a pdf g given by,
'
g(y) = f(x) |dx/dy|
where x is expressed in terms of y. If His increasing, then g is nonzero for
those values of y satisfying H(a) < y < H(b). If H is decreasing, then g
is nonzero for those values of y satisfying H(b) < y < H(a).
Proof: (a) Assume that His a strictly increasing function. Hence
G(y) = P( Y ≤ y) = P(H(X) ≤ y)
= P(X ≤ H-1(y)) = F(H-1(y)).
Differentiating G(y) with respect to y, we obtain, using the chain rule for derivatives,
Thus
dG(y)/dy = dG(y)/dx . dx/dy where x = H-1(y).
G'(y) = dF(x)/dx dx/dy = f(x)dx/dy
(b) Assume that H is a decreasing function. Therefore
G(y) = P(Y ≤ y) = P(H(X) ≤ y) = P(X ≤ H-1(y))
= 1 - P(X ≤ H-1(y)) = 1 - F(H-1(y)).
Proceeding as above, we may write
dG(y)/dy = dG(y)/dx .dx/dy = d[1-F(x)]/dx . dx/dy = -f(x)dx/dy
Note: The algebraic sign obtained in (b) is correct since, if y is a decreasing function of x, x is a decreasing function of y and hence dx/dy < 0. Thus, by using the absolute value sign around dx/dy we may combine the result of (a) and (b) and obtain the final form of the theorem.
EXAMPLE 2
Let us reconsider Examples 5.6 and 5.7, by applying Theorem 1
(a) For Example we had f(x) = 2x, 0 < x < 1, and y = 3x + I. Hence x = (y - 1)/3 and dx/dy = 1/3 Thus g(y) = 2[(y - 1 )/3}1/3 = 4/9(y - 1 ),
1 < y < 4, which checks with the result obtained previously.
(b) In Example 1 we had f(x) = 2x, 0 < x < l , and y = e-x. Hence
x = -in y and dx/dy = - 1 /y. Thus g(y) = -2 (in y)/y, 1/e< y<1 p="" which="">again checks with the above result.
If y = H(x) is not a monotone function of x we cannot apply the above
method directly. Instead, we shall return to the general method outlined above.
The following example illustrates this procedure.
EXAMPLE 3
Suppose that
f(x) = 1/2 -1 < x < 1,
= 0, elsewhere.
Let H(x) = x 2. This is obviously not a monotone function over the interval [ -l, l ] Hence we obtain the pdf of Y = x 2 as follows:
<1 p="" which=""> G(y) = P( Y ≤ y) = P(x 2 ≤ y)
= P( -√y ≤ x≤ √y)
= F(yy) - F(-yy),
where F is the cdf of the random variable X. Therefore
g(y) = G'(y) = f(√y)/2√y - f(-√y)/(-2√y)
<1 p="" which=""> =1[f(√y)' + f(- √y)]/2√y'
<1 p="" which=""> Thus g(y) = (l/2√y) (1/2 + 1/2) = l/2√y, 0 < y < l. (
The method used in the above example yields the following general result.
Theorem 5.2. Let X be a continuous random variable with pdf f Let Y = x 2•
Then the random variable Y has pdf given b
<1 p="" which="">g(y)=<1 p="" which="">1[f(√y)' + f(- √y)]/2√y'
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