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(iii) Differentiate the integral to get that fr(t) 2|t| = 1/2 500 £2² (1u). 2/11 u. fv (u)du, k k where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z2 and U to get that ∞ fr(t) = constant x T (Hint: Z²x²(1).) (v) Apply the change of variable v = (1 + € ) u k+1 u ² + ¹ - ¹e - 1 (¹ + ² ) u du. u to get that fr(t) = constant x k+1 (₁ + ²) - ² k (Hint: everything that doesn't involve t can be dumped into the constant factor.)

(iii) Differentiate the integral to get that fr(t) 2|t| = 1/2 500 £2² (1u). 2/11 u. fv (u)du, k k where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z2 and U to get that ∞ fr(t) = constant x T (Hint: Z²x²(1).) (v) Apply the change of variable v = (1 + € ) u k+1 u ² + ¹ - ¹e - 1 (¹ + ² ) u du. u to get that fr(t) = constant x k+1 (₁ + ²) - ² k (Hint: everything that doesn't involve t can be dumped into the constant factor.)

A First Course in Probability (10th Edition)
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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can you show me the question of 3,4 and 5? Thank you so much for your help

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.
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.
(iii) Differentiate the integral to get that fr(t) = 12 for 12² (2) fz² k 0 fr(t) = constant x where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z2 and U to get that (Hint: Z² ~ x²(1).) (v) Apply the change of variable v = rx 5 (1 + ²/² ) 2|t| k -u. fv(u)du, fr(t) = constant x k+1 u ² + ¹ - ¹ e − ² ( ¹ + ² ) u J ₁ 2 du. u to get that k+1 2 × ( ₁ + ² ) - ** . k (Hint: everything that doesn't involve t can be dumped into the constant factor.)
Transcribed Image Text:(iii) Differentiate the integral to get that fr(t) = 12 for 12² (2) fz² k 0 fr(t) = constant x where fr is the PDF of T and fz2 is the PDF of Z². (iv) Plug in the PDFs of Z2 and U to get that (Hint: Z² ~ x²(1).) (v) Apply the change of variable v = rx 5 (1 + ²/² ) 2|t| k -u. fv(u)du, fr(t) = constant x k+1 u ² + ¹ - ¹ e − ² ( ¹ + ² ) u J ₁ 2 du. u to get that k+1 2 × ( ₁ + ² ) - ** . k (Hint: everything that doesn't involve t can be dumped into the constant factor.)
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