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12.4: Exponential and normal random variables
Exponential density function
Given a positive constant k > 0, the exponential density function (with parameter k) is
f (x) =
ke−kx^ if x ≥ 0 0 if x < 0
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Expected value of an exponential random
variable
Let X be a continuous random variable with an exponential density function with parameter k.
Integrating by parts with u = kx and dv = e−kxdx so that du = kdx and v = − k^1 e−kx, we find
E(X) =
−∞
xf (x)dx
=
0
kxe−kxdx
= (^) r→liminf ty[−xe−kx^ − 1 k
e−kx]|r 0
= (^) k^1
Variance of exponential random variables
Integrating by parts with u = kx^2 and dv = e−kxdx so that du = 2kxdx and v = − k^1 e−kx, we have
∫ (^) ∞
0
x^2 e−kxdx = (^) rlim→∞([−x^2 e−kx]|r 0 + 2
∫ (^) r
0
xe−kxdx)
= (^) rlim→∞([−x^2 e−kx^ − 2 k
xe−kx^ − 2 k^2
e−kx]|r 0 )
= (^) k^22
So, Var(X) = (^) k^22 − E(X)^2 = (^) k^22 − (^) k^12 = (^) k^12.
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Example
Exponential random variables (sometimes) give good models for the time to failure of mechanical devices. For example, we might measure the number of miles traveled by a given car before its transmission ceases to function. Suppose that this distribution is governed by the exponential distribution with mean 100, 000. What is the probability that a car’s transmission will fail during its first 50 , 000 miles of operation?
Taylor expansion for the normal cumulative
distribution function
Let f (x) = √^12 π e
− 21 x 2 be the standard normal density function and let F (x) =
∫ (^) x −∞ f^ (t)dt^ be the standard normal^ cumulative distribution function.
We compute a Taylor series expansion,
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G(x) =
2 π
e −^21 x^2 dx
= √^1
2 π
n=
(−1)n n!2n^ x
2 ndx
= √^1
2 π
∑^ ∞
n=
(−1)n n!2n(2n + 1) x
2 n+
= √^1
2 π
(x − 16 x^3 + 401 x^5 − 3361 x^7 + · · · )
So F (x) = G(x) + C for some C. As 0 is the expected value, we need 12 = F (0) = G(0) + C = C.
Example
If the continuous random variable X is normally distributed, what is the probability that it takes on a value of more than a standard deviations above the mean?
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Solution
Via a change of variables, we may suppose that X is normally distributed with respect to the standard normal distribution. Let F be the cumulative distribution function for the standard normal distribution.
Pr(X ≥ 1) =
1
√^1
2 π
e
− 21 x 2 dx = (^) rlim→∞(F (r) − F (1))
≈ (^) rlim→∞(F (r)) − (^12 + √^1 2 π
3 6 +
40 )^ −^
