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1 Probability Density Functions
(a) What are the two properties of a probability density function f (x)?
(i) f (x) ≥ 0 for all x ∈ (−∞, ∞)
(ii)
−∞
f (x)dx = 1
(b) Which of the following are valid probability density functions? Explain why or why not, a yes or no is not a sufficient answer.
f (x) =
x^2 + 2x, for − 1 ≤ x ≤ 1 0 , otherwise.
g(x) =
2 x, for 0 ≤ x ≤ 1 0 , otherwise.
We need to check if the functions satisfy the two criteria in (a). Because f (−1) = − 1 < − 0 , f (x) fails to satisfy property (i). Therefore, f (x) is not a valid p.d.f. On the other hand, g(x) ≥ 0 for all x ∈ (−∞, ∞). Also ∫ (^) ∞
−∞
g(x)dx =
0
2 xdx = x^2
^1
Since g(x) satisfies the two properties, g(x) is a valid p.d.f. By a result stated in class, there is a r.v. with p.d.f. g(x).
(c) Find a constant C such that the function
h(x) =
Cx^3 (1 − x), for 0 ≤ x ≤ 1 0 , otherwise
is a probability density function. Let X be a random variable with probability density function h(x). What is E[X]? What is V ar(X)? ∫ (^1)
0
Cx^3 (1 − x) =
C
x^4
∣^1
0 −^
C
x^5
∣^1
0 =^
C
set = 1 iff C = 20
E[X] =
−∞
xh(x)dx =
0
20 x^4 (1 − x) =
E[X^2 ] =
−∞
x^2 h(x)dx =
0
20 x^5 (1 − x) =
V ar(X) = E[X^2 ] − (E[X])^2 =
2 Cumulative Distribution Functions
(a) State the three properties of a cumulative distribution function F (x).
(i) F(x) is non-decreasing (ii) F(x) is right-continuous (iii) lim x→−∞ F (x) = 0, and lim x→∞ F (x) = 1
(b) Find expected value and variance of the random variable X with cumulative distribution function
FX (t) =
0 for t < 0 1 − e−^3 t^ for t ≥ 0
Also, find P (X ≤ 5) and P (X > 7). First, differentiate FX (t) with respect to t (at points where FX (t) is differentiable) to find the p.d.f. fX (t) corresponding to FX (t):
fX (t) =
{ (^) dF X (t) dt = 3e
− 3 t, for 0 < x 0 , otherwise
Now, use the definition of the expected value of a continuous random variable to compute E[X].
E[X] =
−∞
xfX (x)dx =
0
3 xe−^3 x^ = 1/ 3.
Integration by parts (
udv = uv −
vdu), with dv = e−^3 xdx and u = 3x can be used to compute the integral above. If you’re rusty on integration by parts, check out your old calculus notes, or (better yet), come ask me! To compute V ar(X), we can use the formula V ar(X) = E[X^2 ] − (E[X])^2. Another application of integration by parts (again, if this integration by parts business is confusing you, come ask me) gives
E[X^2 ] =
0
3 x^2 e−^3 xdx = 2/ 9
V ar(X) = 2 / 9 − 1 /9 = 1/ 9
Finally,
P (X ≤ .5) = FX (.5) = 1 − e−^1.^5 = 0. 78 P (X > 1 /3) = 1 − FX (1/3) = e−^1 = 0. 37
(c) Find the c.d.f of the random variable X with p.d.f. h(x) given in part (c) of Problem 2 above. What is P (X ≤ .5)? The c.d.f. of X is
H(x) =
∫ (^) x
0
h(t)dt =
0 for x < 0 , 5 x^4 − 4 x^5 for 0 ≤ x < 1 , 1 for x ≥ 1
Using the c.d.f,
P (X ≤ .5) = H(.5) = 5/ 16 − 4 /32 = 0. 1875.
(c) Do you recognize the distribution of Z? State expected value and variance. Interpret the expected value. E[Z] = 1/ 0 .45 = 2. 22 , V ar[Z] = 1/ 0. 452 = 4. 94 on average the contest is over in 2.22 hours.
