(2) Prove that the volume under the bivariate normal density is one. [Hint: by a substitution eliminate the constants a¡ , a2, b1 , bz from the problem.] Here is the proof. Justify the equalities at the marked spots. By the change of variable u = (x - a1 Vb, and v = (y - az)/bz , no loss of generality takes place if we assume that a = az = 0 and that bị = b2 = 1. Take 2xV1- p = (1/C). Then we have -1 c exp{ 2 * + y? – 2pxy) } dy = c / exp{ - 2pxy + p°x² +G² , dy 2(1 – p2) -1 { 2(1 – 2 1 V2a(1 – p°) = exp -(y - px) dy 3D Then, (iii), explain why does the prove derivation prove that the volume under the bivariate normal density is one. The correct answers for the three parts are: Taylor expansion Derivative in y Integration by parts Pulled ex12 out of integral None of the above N/A (i- Select One) Taylor expansion Derivative in y Integrating a normal density Integration by parts None of the above N/A (ii- Select One) Integrating a density gives 1 Taylor expansion Derivative in y Integration by parts None of the above N/A (iii- Select One)
(2) Prove that the volume under the bivariate normal density is one. [Hint: by a substitution eliminate the constants a¡ , a2, b1 , bz from the problem.] Here is the proof. Justify the equalities at the marked spots. By the change of variable u = (x - a1 Vb, and v = (y - az)/bz , no loss of generality takes place if we assume that a = az = 0 and that bị = b2 = 1. Take 2xV1- p = (1/C). Then we have -1 c exp{ 2 * + y? – 2pxy) } dy = c / exp{ - 2pxy + p°x² +G² , dy 2(1 – p2) -1 { 2(1 – 2 1 V2a(1 – p°) = exp -(y - px) dy 3D Then, (iii), explain why does the prove derivation prove that the volume under the bivariate normal density is one. The correct answers for the three parts are: Taylor expansion Derivative in y Integration by parts Pulled ex12 out of integral None of the above N/A (i- Select One) Taylor expansion Derivative in y Integrating a normal density Integration by parts None of the above N/A (ii- Select One) Integrating a density gives 1 Taylor expansion Derivative in y Integration by parts None of the above N/A (iii- Select One)
MATLAB: An Introduction with Applications
6th Edition
ISBN:9781119256830
Author:Amos Gilat
Publisher:Amos Gilat
Chapter1: Starting With Matlab
Section: Chapter Questions
Problem 1P
Related questions
Question
![**Problem Statement:**
(2) Prove that the volume under the bivariate normal density is one. [Hint: by a substitution eliminate the constants \( a_1, a_2, b_1, b_2 \) from the problem.]
**Proof Explanation:**
Here is the proof. Justify the equalities at the marked spots.
By the change of variable \( u = (x-a_1)/b_1 \) and \( v = (y-a_2)/b_2 \), no loss of generality takes place if we assume that \( a_1 = a_2 = 0 \) and that \( b_1 = b_2 = 1 \). Take \( 2\pi \sqrt{1-\rho^2} = (1/C) \). Then we have
\[ C \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (x^2+y^2-2\rho xy) \right\} dy = C \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (y^2 - 2\rho xy + \rho^2 x^2 + (x^2 - \rho^2 x^2)) \right\} dy \]
\[ (i) = C e^{-x^2/2} \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (y-\rho x)^2 \right\} dy \]
\[ (ii) = C e^{-x^2/2} \sqrt{2\pi(1-\rho^2)} = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \]
**Question:**
Then, (iii), explain why does the prove derivation prove that the volume under the bivariate normal density is one.
**Multiple Choice Answers:**
The correct answers for the three parts are:
- **(i- Select One):**
- [ ] Taylor expansion
- [ ] Derivative in \( y \)
- [ ] Integration by parts
- [ ] Pulled \( e^{-x^2/2} \) out of integral](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F421db798-cca4-4efb-a6a9-3775d86708c7%2F7d6fcf1e-b46e-4fce-b855-4308cf801f14%2Fxw2msgs_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
(2) Prove that the volume under the bivariate normal density is one. [Hint: by a substitution eliminate the constants \( a_1, a_2, b_1, b_2 \) from the problem.]
**Proof Explanation:**
Here is the proof. Justify the equalities at the marked spots.
By the change of variable \( u = (x-a_1)/b_1 \) and \( v = (y-a_2)/b_2 \), no loss of generality takes place if we assume that \( a_1 = a_2 = 0 \) and that \( b_1 = b_2 = 1 \). Take \( 2\pi \sqrt{1-\rho^2} = (1/C) \). Then we have
\[ C \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (x^2+y^2-2\rho xy) \right\} dy = C \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (y^2 - 2\rho xy + \rho^2 x^2 + (x^2 - \rho^2 x^2)) \right\} dy \]
\[ (i) = C e^{-x^2/2} \int_{-\infty}^{\infty} \exp \left\{ \frac{-1}{2(1-\rho^2)} (y-\rho x)^2 \right\} dy \]
\[ (ii) = C e^{-x^2/2} \sqrt{2\pi(1-\rho^2)} = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \]
**Question:**
Then, (iii), explain why does the prove derivation prove that the volume under the bivariate normal density is one.
**Multiple Choice Answers:**
The correct answers for the three parts are:
- **(i- Select One):**
- [ ] Taylor expansion
- [ ] Derivative in \( y \)
- [ ] Integration by parts
- [ ] Pulled \( e^{-x^2/2} \) out of integral
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