Choose the best justification of the two steps (i) and (ii). Newton's law Triangle inequality Partial fractions Integration by parts Definition of cdf N/A (Step (i) because of) Newton's law Triangle inequality Partial fraction Integration by parts Definition of cdf NIA (Step (ii) because of) O

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Author:Amos Gilat
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Chapter1: Starting With Matlab
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(13) Let X have a chi square density with n = 4 degrees of freedom. Show that its cdf is:
(t/2)
F() %3D 1 — еxp(-i/2) >
t > 0.
j=0
Here is the proof. For any real number t > 0,
(1/2)?
F(t) =
ex/2 dx,
„2–1
(i)
T(2)
1
-2te-i/2
4
+ 2
dx
(ii)
(-2te-2 + 4(1 – e 2))
–t/2
= 1 - e-12 – (t/2)e¬t2.
Choose the best justification of the two steps (i) and (ii).
Newton's law
Triangle inequality
Partial fractions
Integration by parts
Definition of cdf
N/A
(Step (i) because of)
Newton's law Triangle inequality Partial fraction Integration by parts Definition of cdf N/A
(Step (ii) because of) O
Transcribed Image Text:(13) Let X have a chi square density with n = 4 degrees of freedom. Show that its cdf is: (t/2) F() %3D 1 — еxp(-i/2) > t > 0. j=0 Here is the proof. For any real number t > 0, (1/2)? F(t) = ex/2 dx, „2–1 (i) T(2) 1 -2te-i/2 4 + 2 dx (ii) (-2te-2 + 4(1 – e 2)) –t/2 = 1 - e-12 – (t/2)e¬t2. Choose the best justification of the two steps (i) and (ii). Newton's law Triangle inequality Partial fractions Integration by parts Definition of cdf N/A (Step (i) because of) Newton's law Triangle inequality Partial fraction Integration by parts Definition of cdf N/A (Step (ii) because of) O
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