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The concepts of expected value and variance for a continuous random variable. It provides formulas for calculating these statistics and interprets their meanings. The document also includes examples of calculating expected value and variance for uniform, exponential, and normal distributions.
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If X is a random variable with corresponding probability density function f (x), then we define the expected value of X to be
−∞
xf (x)dx
We define the variance of X to be
Var(X) :=
−∞
[x − E(X)]^2 f (x)dx
1
As with the variance of a discrete random variable, there is a simpler formula for the variance.
Var(X) =
−∞
[x − E(X)]f (x)dx
=
−∞
[x^2 − 2 xE(X) + E(X)^2 ]f (x)dx
=
−∞
x^2 f (x)dx − 2 E(X)
−∞
xf (x)dx
−∞
f (x)dx
=
−∞
x^2 f (x)dx − 2 E(X)E(X) + E(X)^2 × 1
=
−∞
x^2 f (x)dx − E(X)^2
3
The expected value should be regarded as the average value. When X is a discrete random variable, then the expected value of X is precisely the mean of the corresponding data.
The variance should be regarded as (something like) the average of the difference of the actual values from the average. A larger variance indicates a wider spread of values.
As with discrete random variables, sometimes one uses the standard deviation, σ =
Var(X), to measure the spread of the distribution instead.
We compute ∫ (^) ∞
−∞
x^2 f (x)dx =
0
x^2 dx
= 13 x^3 |x x=1=
= (^13)
7
Hence,
Var(X) =
−∞
x^2 f (x)dx − E(X)^2
= 13 − (^14)
= 121
Let X be the random variable with probability density function
f (x) =
ex^ if x ≤ 0 0 if x > 0
Compute E(X) and Var(X).
9
Integrating by parts with ∫ u = x and dv = exdx, we see that xexdx = xex^ − ex^ + C. Thus,
−∞
xf (x)dx
−∞
xexdx
0 r→−∞^ lim
r
xexdx = (^) r→−∞lim [− 1 − rer^ + er^ ] = 1
[We used L’Hˆopital’s rule to see that limr→−∞ rer^ = limr→−∞ (^) e−rr = limr→−∞ (^) −e^1 −r = 0.]
First, we must find the probability density function of X. Differentiating we find that the function
f (x) =
cos(x) if 0 ≤ x ≤ π 2 0 otherwise
is the derivative of F at all but two points. Thus, f (x) is a probability density function for X.
13
−∞
xf (x)dx
∫ π 2
0
x cos(x)dx
= (x sin(x) + cos(x))|x=^
π 2 x= = π 2 − 1
Integrating by parts, we compute
Var(X) =
∫ π 2
0
x^2 cos(x)dx − E(X)^2
= (x^2 sin(x) − 2 sin(x) + 2x cos(x))|x=^
π 2 x=0 −^ (^
π 2
= π
2 4 −^2 −^ (^
π^2 4 −^ π^ + 1) = π − 3
