Let U~ x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be 1 fv(u) = /u³-¹e-³, u > 0
Let U~ x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived the form of the PDF of U to be 1 fv(u) = /u³-¹e-³, u > 0
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
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Please show me the question of I and II, thank you so much
Transcribed Image Text:Let U ~ x²(k), a chi-squared distribution with k degrees of freedom. In lecture, we derived
the form of the PDF of U to be
k
fv(u) = 1/2 u 2 e-²₂ u > 0
where is the normalization constant making this function a PDF. Now, let Z ~ N(0, 1)
be independent from U, and define
Fr(t)
=
T
Then T~ t(k), a t distribution with k degrees of freedom. In this exercise, you will derive
the form of the PDF of T following the same type of calculations in lecture.
(i) Let FT be the CDF of the random variable T. Verify that
=
12
Z
√U/k
{1+P
+ P(Z² ≤ U)
¹P (Z² ≤ U)
ift > 0,
ift < 0.
(Hint: consider the event |Z| ≤ |t|√U/k and draw a picture of the PDF of Z.)
(ii) Use the Law of Total Probability to show that
P ( 2² ≤ ²U) = [° F₂² (u) fv(u)du,
<
k
where Fz2 is the CDF of Z2 and fu is the PDF of U.
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