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FINM7403 Bond Part
Creation date：2024-05-23 15:16:04
Belonging category：金融Finance
Bond Part

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FINM7403 Bond Part II Tutorial Solutions
BKM Chapter 15 problem sets, Q4, 5, 6, 9*, 13, 15, 16*.
Question 4
Suggested Solutions:
The liquidity theory holds that investors demand a premium to compensate them for interest rate exposure and
the premium increases with maturity. Add this premium to a flat curve and the result is an upward sloping yield
curve.
Question 5
Suggested Solutions:
The pure expectations theory, also referred to as the unbiased expectations theory, purports that forward rates
are solely a function of expected future spot rates. Under the pure expectations theory, a yield curve that is
upward (downward) sloping, means that short-term rates are expected to rise (fall). A flat yield curve implies
that the market expects short-term rates to remain constant.
Question 6
Suggested Solutions:
The yield curve slopes upward because short-term rates are lower than long-term rates. Since market rates are
determined by supply and demand, it follows that investors (demand side) expect rates to be higher in the future
than in the near-term.

Question 9
Suggested Solutions:
First calculate the short rates
1
2
2
3
3
4
4
( ) 5.00%
1.06( ) 7.01%
1.05
1.065( ) 7.51%
1.05 1.0701
1.07( ) 8.51%
1.05 1.0701 1.0751
E r
E r
E r
E r
=
= =
= =
×
= =
× ×
In one year, the market’s expectation of the yield curve is:
Bond Years to Maturity YTM(%)
B 1 7.01%
C 2 7.26%
D 3 7.67%

Question 13
Suggested Solutions:
Year
Forward
Rate
PV of $1 received at period end
1 5% $1/1.05 = $0.9524
2 7 1/(1.05×1.07) = $0.8901
3 8 1/(1.05×1.07×1.08) = $0.8241
a. Price = ($60 × 0.9524) + ($60 × 0.8901) + ($1,060 × 0.8241) = $984.14
b. To find the yield to maturity, solve for y in the following equation:
$984.10 = [$60 × Annuity factor (y, 3)] + [$1,000 × PV factor (y, 3)]
This can be solved using a financial calculator to show that y = 6.60%:
PV = -$984.10; N = 3; FV = $1,000; PMT = $60. Solve for I = 6.60%.
c.
Period Payment Received at End of Period: Will Grow by a Factor of: To a Future Value of:
1 $60.00 1.07 × 1.08 $69.34
2 60.00 1.08 64.80
3 1,060.00 1.00 1,060.00
$1,194.14
$984.10 × (1 + y realized)3 = $1,194.14
1 + y realized = 0666.1
10.984$
14.194,1$ 3/1
=




 ⇒ y realized = 6.66%
Alternatively, PV = -$984.10; N = 3; FV = $1,194.14; PMT = $0. Solve for I = 6.66%.

d. Next year, the price of the bond will be:
[$60 × Annuity factor (7%, 2)] + [$1,000 × PV factor (7%, 2)] = $981.92
Therefore, there will be a capital loss equal to: $984.10 – $981.92 = $2.18
The holding period return is: %88.50588.0
10.984$
)18.2$(60$
==
−+
Question 15
Suggested Solutions:
The price of the coupon bond, based on its yield to maturity, is:
[$120 × Annuity factor (5.8%, 2)] + [$1,000 × PV factor (5.8%, 2)] = $1,113.99
If the coupons were stripped and sold separately as zeros, then, based on the yield to maturity of zeros with
maturities of one and two years, respectively, the coupon payments could be sold separately for:
08.111,1$
06.1
120,1$
05.1
120$
2 =+
The arbitrage strategy is to buy zeros with face values of $120 and $1,120, and respective maturities of one year
and two years, and simultaneously sell the coupon bond. The profit equals $2.91 on each bond.
Question 16
Suggested Solutions:
a. The one-year zero-coupon bond has a yield to maturity of 6%, as shown below:
1
$100$94.34
1 y
=
+
⇒y1 = 0.06000 = 6.000%
The yield on the two-year zero is 8.472%, as shown below:
2
2
$100$84.99
(1 )y
=
+
⇒y2 = 0.08472 = 8.472%
b. The price of the coupon bond is: 51.106$
)08472.1(
112$
06.1
12$
2 =+

Therefore: yield to maturity for the coupon bond = 8.333%
[On a financial calculator, enter: n = 2; PV = –106.51; FV = 100; PMT = 12]
c.
2 2
2
2
1
(1 ) (1.08472)1 1 0.1100 11.00%
1 1.06
yf
y
+
= − = − = =
+
d. Expected price =
$112
1.11
=$100.90
(Note that next year, the coupon bond will have one payment left.)
Expected holding period return = %00.60600.0
51.106$
)51.106$90.100($12$
==
−+
This holding period return is the same as the return on the one-year zero.
e. If there is a liquidity premium, then: E(r2) < f 2
E(Price) =
2
$112 $100.90
1 ( )E r
>
+
E(HPR) > 6%

CFA Problems 2, 3, and 5*.
Problem 2
Suggested Solutions:
d. Investors bid up the price of short term securities and force yields to be relatively low, while doing just the
opposite at the long end of the term structure. Therefore, they must be compensated more for giving up liquidity
in the long term.
Problem 3
Suggested Solutions:
a. (1+y4 )4 = (1+ y3 )3 (1 + f 4 )
(1.055)4 = (1.05)3 (1 + f 4 )
1.2388 = 1.1576 (1 + f 4 ) ⇒ f 4 = 0.0701 = 7.01%
b. The conditions would be those that underlie the expectations theory of the term structure: risk neutral market
participants who are willing to substitute among maturities solely on the basis of yield differentials. This
behaviour would rule out liquidity or term premia relating to risk.
c. Under the expectations hypothesis, lower implied forward rates would indicate lower expected future spot
rates for the corresponding period. Since the lower expected future rates embodied in the term structure are
nominal rates, either lower expected future real rates or lower expected future inflation rates would be consistent
with the specified change in the observed (implied) forward rate.

Problem 5
Suggested Solutions:
The present value of each bond’s payments can be derived by discounting each cash flow by the appropriate rate
from the spot interest rate (i.e., the pure yield) curve:
Bond A: 53.98$
11.1
110$
08.1
10$
05.1
10$PV 32 =++=
Bond B: 36.88$
11.1
106$
08.1
6$
05.1
6$PV 32 =++=
Bond A sells for $0.13 (i.e., 0.13% of par value) less than the present value of its stripped payments. Bond B
sells for $0.02 less than the present value of its stripped payments. Bond A is more attractively priced.

BKM Chapter 16 problem sets, Q1, 2, 3, 4*, 5, 7, 9, 12*, 14*.
Question 1
Suggested Solutions:
While it is true that short-term rates are more volatile than long-term rates, the longer duration of the longer-
term bonds makes their prices and their rates of return more volatile. The higher duration magnifies the
sensitivity to interest-rate changes.
Question 2
Suggested Solutions:
Duration can be thought of as a weighted average of the maturities of the cash flows paid to holders of the
perpetuity, where the weight for each cash flow is equal to the present value of that cash flow divided by the
total present value of all cash flows. For cash flows in the distant future, present value approaches zero (i.e., the
weight becomes very small) so that these distant cash flows have little impact and, eventually, virtually no
impact on the weighted average.
Question 3
Suggested Solutions:
The percentage change in the bond’s price is:
7.194 0.005 0.0327 3.27%,
1 1.10
D y
y
− ×∆ = − × = − = −
+
or a 3.27% decline
Question 4
Suggested Solutions:
a. YTM = 6%
(1) (2) (3) (4) (5)
Time until Payment
(Years)
Cash Flow PV of CF (Discount Rate = 6%) Weight Column (1) × Column (4)
1 60.00 56.60 0.0566 0.0566

2 60.00 53.40 0.0534 0.1068
3 1,060.00 890.00 0.8900 2.6700
Column sums $1,000.00 1.0000 2.8334
Duration = 2.833 years
b. YTM = 10%
(1) (2) (3) (4) (5)
Time until
Payment (Years)
Cash Flow PV of CF (Discount Rate = 10%) Weight Column (1) × Column (4)
1 60.00 54.55 0.0606 0.0606
2 60.00 49.59 0.0551 0.1102
3 1,060.00 796.39 0.8844 2.6532
Column sums $900.53 1.0000 2.8240
Duration = 2.824 years, which is less than the duration at the YTM of 6%.
Question 5
Suggested Solutions:
For a semi-annual 6% coupon bond selling at par, we use the following parameters: coupon = 3% per half-year period,
y = 3%, T = 6 semi-annual periods.
(1) (2) (3) (4) (5)
Time until Payment
(Years)
Cash Flow PV of CF (Discount Rate = 3%) Weight Column (1) × Column (4)
1 $ 30.00 $ 29.13 0.02913 0.02913
2 30.00 28.28 0.02828 0.05656
3
30.00
27.45 0.02745 0.08236
4 30.00 26.65 0.02665 0.10662
5 30.00 25.88 0.02588 0.12939
6 1030.00 862.61 0.86261 5.17565
Column sums $1000.00 1.00000 5.57971
D = 5.5797 half-year periods = 2.7899 years
If the bond’s yield is 10%, use a semi-annual yield of 5% and semi-annual coupon of 3%:
(1) (2) (3) (4) (5)
Time until Payment
(Years)
Cash Flow PV of CF (Discount Rate = 5%) Weight Column (1) × Column (4)
1 $ 30.00 $ 28.57 0.03180 0.03180
2 30.00 27.21 0.03029 0.06057

3
30.00
25.92 0.02884 0.08653
4 30.00 24.68 0.02747 0.10988
5 30.00 23.51 0.02616 0.13081
6 1030.00 768.60 0.85544 5.13265
Column sums $898.49 1.00000 5.55223
D = 5.5522 half-year periods = 2.7761 years
Question 7
Suggested Solutions:
D. Investors tend to purchase longer term bonds when they expect yields to fall so they can capture significant
capital gains, and the lack of a coupon payment ensures the capital gain will be even greater.
Question 9
Suggested Solutions:
a.
(1) (2) (3) (4) (5)
Time until Payment
(Years)
Cash Flow PV of CF (Discount Rate = 10%) Weight Column (1) × Column (4)
1 $10 million $ 9.09 million 0.7857 0.7857
5 4 million 2.48 million 0.2143 1.0715
Column sums $11.57 million 1.0000 1.8572
D = 1.8572 years = required maturity of zero-coupon bond.
b. The market value of the zero must be $11.57 million, the same as the market value of the obligations.
Therefore, the face value must be: $11.57 million × (1.10)1.8572 = $13.81 million

Question 12
Suggested Solutions:
a. PV of the obligation = $10,000 × Annuity factor (8%, 2) = $17,832.65
(1) (2) (3) (4) (5)
Time until Payment
(Years)
Cash Flow PV of CF (Discount Rate = 8%) Weight Column (1) × Column (4)
1 $10,000.00 9,259.259 0.51923 0.51923
2 10,000.00 8,573.388 0.48077 0.96154
Column sums $17,832.647 1.00000 1.48077
D = 1.4808 years
b. A zero-coupon bond maturing in 1.4808 years would immunize the obligation. Since the present value of the
zero-coupon bond must be $17,832.65, the face value (i.e., the future redemption value) must be
$17,832.65 × 1.081.4808 = $19,985.26
c. If the interest rate increases to 9%, the zero-coupon bond would decrease in value to
92.590,17$
09.1
26.985,19$
4808.1 =
The present value of the tuition obligation would decrease to $17,591.11
The net position decreases in value by $0.19
d. If the interest rate decreases to 7%, the zero-coupon bond would increase in value to
99.079,18$
07.1
26.985,19$
4808.1 =
The present value of the tuition obligation would increase to $18,080.18
The net position decreases in value by $0.19
The reason the net position changes at all is that, as the interest rate changes, so does the duration of the stream
of tuition payments.

Question 14
Suggested Solutions:
a. The duration of the perpetuity is: 1.05/0.05 = 21 years
Call w the weight of the zero-coupon bond. Then
(w × 5) + [(1 – w) × 21] = 10 ⇒ w = 11/16 = 0.6875
Therefore, the portfolio weights would be as follows: 11/16 invested in the zero and 5/16 in the perpetuity.
b. Next year, the zero-coupon bond will have a duration of 4 years and the perpetuity will still have a 21-year
duration. To obtain the target duration of nine years, which is now the duration of the obligation, we again solve
for w:
(w × 4) + [(1 – w) × 21] = 9 ⇒ w = 12/17 = 0.7059
So, the proportion of the portfolio invested in the zero increases to 12/17 and the proportion invested in the
perpetuity falls to 5/17.

CFA Problems 2, 3*, 4a.i. 4.a.ii. only.
Problem 2
Suggested Solutions:
a. Bond price decreases by $80.00, calculated as follows:
10 × 0.01 × 800 = 80.00
b. ½ × 120 × (0.015)2 = 0.0135 = 1.35%
c. 9/1.10 = 8.18
d. (i)
e. (i)
f. (iii)

Problem 3
Suggested Solutions:
a. Modified duration 26.9
08.1
10
YTM1
durationMacaulay
==
+
= years
b. For option-free coupon bonds, modified duration is a better measure of the bond’s sensitivity to changes in
interest rates. Maturity considers only the final cash flow, while modified duration includes other factors, such
as the size and timing of coupon payments, and the level of interest rates (yield to maturity). Modified duration
indicates the approximate percentage change in the bond price for a given change in yield to maturity.
c. i. Modified duration increases as the coupon decreases.
ii. Modified duration decreases as maturity decreases.
d. Convexity measures the curvature of the bond’s price-yield curve. Such curvature means that the duration
rule for bond price change (which is based only on the slope of the curve at the original yield) is only an
approximation. Adding a term to account for the convexity of the bond increases the accuracy of the
approximation. That convexity adjustment is the last term in the following equation:
* 21( ) Convexity ( )
2
P D y y
P
∆  = − ×∆ + × × ∆  
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CSCI262 SCC.311 COMP8101 IAB320 CHC5028 TSP ICC565 CS512 320SC COMP2396 COMP207 SVM BSCH EECS 2101 COMP3230 CEG3136 INFO 3440 EE3009 MGT7174 COMP 3334 ECON20120 Econ 702 CFRM 505 CS4001 CSE 328 AMATH 482 ENG2008 MATH39522 DTS101TC CSC318 EDUC 124 NBA 5110 ECON0125 ELEC7310 EEP101 ECE 777AE MDS 6112 Bioc0036 ACCFIN5246 STA 404 CFRM505 Chinese 2241G MATH 376 AUENV 218 ELEC0006 ACCT 3230 FIN3020 ENVM3115 CMPSC 464 EE 4233 CST 8284 ENGI 2211 EEB240 COMP52715 PHYS1120 ECON0019 MAT187 FINN1037 CSci 5421 7SSMM803 STAT 400 LUBS5996M ELC5216 STATS 4A03 ENGG1003 BADM 3661 DATA423 ST332 ECO2121 6406ELE IB3K20 CSE 465 CSE115 CMPUT 206 ecs150 INU1111 QM222 DTS204TC MATH2319 CPT108 SENG6110 MATH3041 INFS3604 MA 4704 CIVIL 750 EEEE4123 CS 3800 comp3511 MEC206 CSCI-UA.202 IOM 302 NBS8295 BMS 201 ECON 3018 ESE 605 ADM3225 STAT7038 Economics 409 ECON 503 MSE800 4SSMN902 IEOR E4106 FRE6831 ISE 563 MATH 451 ECON0001 COMPSCI369 CITS4402 MA1608 MQBS1030 MUSC1982 CITS4012 ACC202 BAFI1026 MA117 ECON5074 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270M CIS 3207 Programs CSCI 2132 CS270 CSE-381 COMP3230A CIS 212 CSI 333 FINM 326 FINM-36702 Introduction to Software Analysis EEE102 GNG1106 ECS 60 CS 161 CSCI851 CSE1222 EDPS0249 TCHR5003 ANTA02 BBUS1SBY SOSC 1140 MDIA2091 FINS3641 INSY 661 ITEC3040 CPSC 481 computer DEE3519 COMP304 CSE536 Microtheory OFFICE 280 CIVE50003 COSC2673 ECS784U CMPSC/MATH 455 AMA 505 CSE 158 BUS1040 FIT3179 FIT5202 Math 104A Math 111A MBUS107 COMPSCI 761 COMP3710 COMPSCI 316 COMP559 COMP557 CP1 1902 COMP 310 ECED 3403 COMPX523 AGCM 3103 EMET 7001 ENGR3821 CISC 124 IPC comp20005 CS 323 CS105 STAT 326 MATH4007 CS4551 COMP1000 ACIT 4640 ENVX3002 CS2034 FC712 algorithms CSCI 4155 CSI4142 POL 850 MATH20811 2B03 CSCI803 Economies COMP 2016 Architect TRADE INFO1110 FIT1043 networks Robotics ICS 31 ECEC-301 FIT2099 CS 3640 CSE1010 Computational COMP1001 CSE 231 CS10 FIT5166 CSE 400 CSCI561 Math 350 CO-496 ENGG6400 CSCI 1133 COMP3055 MATH 819 CSE 6363 CIS 555 CS4420 COMP0037 ECS 140A Spanning Tree Protocol network 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4061 ECON 8820 INT104 ITE3713 CS551J ECM1410 CSMBD21 COMP 2012 CISC221 Code SCC312 COMP639 CMPSC 221 system COMP3217 COMP1721 ENG5056 Math 466 COMP0039 MFIN 5400F FINN40915 EENG31400 MODL5007M ​EEET 3050 EEET 3050 Quantitative Methods mathematics COMP1921 FIN3309 VE311 ESCI 1908 FIT5057 COSC2446 COMPUTING Math 110A EE 567 MATH10098 CISC642 EE10134 ECE 321 MTH2051 SIT 281 EEE3314 ELEC273 COMP5930M CE322 ELEC270 EE5101 MCEN90008 MTSC 887 Probability ENGN4528 MAE 424 MA2552 EBU5303 CHEM10600 EBU6018 ELEC2103 ELEE 5200 CSE 535 COMP6528 IS6153 IS3S664 BUSANA 7003 QTS0109 ECO3152B ENGR 227 LCSCI4208 MATH32051 ECS776P MGMT3004 N1551 JAVA ROB-6213 TNS3113 BEEM062 COMP47670 4SSMN903 BST 833 COEN 146 BASC0003 CAS EC501 Networks CSIT940 ELEC372 ACCFIN5034 Math 3527 DATA4000 INT102 FND109 MMM071 ROB6213 ETF5650 ECON 445 QMSS 301 Econ 320 PHY 4390 FIN 111 FINN41615 ECS 116 MTH6158 IOE 552 STA 141B Math 413 ECON3016 Physics 4303 Numerical BFF5270 EECS 111 FINANCE 251 COMP814 MGSC 410 ECON0051 EBU5376 STAT6118 ECON 20110 COMU7000 average acceleration BUSI4412 BFF3121 BIT233 ECMM462 ETC3430 DESN2348 INTE2584 ENME502 MA3XJ INTE2627 MA2815 MA3AM L0101 Math 3MB3 CPCCBC4004 EEEM061 0502E220 APPLIED 6SSMN310 ENV 3310 PM2PCOL3 CONS6201 MATH502 FIT5003 EEEN313 MATH3001 MECH3460 ECON30009 COSC1252 Assembler COMP 3023 ENGG7601 COMP20005 COMP SCI 1103 ICSI 402 INF2C CS320 COMPSCI7064 ECE 3220 PROGRAMMAMING MIPS Computer Science 2413 AM3611 MSOA TE2101 FFT CMPSC-121 CSE 109 COMP SCI 7064 HCI CMP-5013A CCN2042 COMP3400 EEEN30052 CO4215 EE101 KIT107 CSE 421 CSE521 MCD4720 EECS 280 MIT 403 GRPC VG101 APiE EECE 1080C FIT3143 CSE 325 ISOM 3029 CS3307 State SIT255 Engineering MQF Modelling ACX3150 SIA322 ACCT90004 COMU7201 MGMT90018 WRIT1000 COMP3702 BISM1201 COMP7505 IDEA9106 CSE3SMT BFC5926 BFC5280 BFC5935 QBUS3330 ECON3700 IFN521 ECON2121 COSC2626 MA413 DDES9903 ENGG1300 COMP9337 CO2201 ACCT2002 ICM289 MMM054 MSBA-204 DIGM707 CSC2250 Econ312 EVAT python CAN404 LUBS2840 CSCI-GA.2250 COMSM0093 COMP281 AFM 333 IBUS1001 CP2501 Math 105A VE320 TEC101 SOCY2166 COM240 EFIM20033 ENGI4467 EFIM20034 EC 979 BE432 7EJ502 CSSE6400 COMP90025 Econ 312 ENG2031 ECO0006 EXERSCI 207 CAN209 ECO 137W FT312 ECO 2199 MATH32052 BDW4213 APS 360 SAS 6 CHME0017 PO92Q Economics 01AV MATH39512 MATH 5486 COMP8620 ISE102 ECON235 ECON 206 FI 345 PHAY0020 MATH0099 MATH0085 EFB106 PHY 134 MAE115 circuits Chemistry CSE12 MATH-UA 233 PHAS0038 COMP2140 CONMGNT 7049 COMP151101 ECON 485 ACX5903 ENGN6536 ARCH1162 Law PHAS0042 CCF1C03 MATH 351 Phys 139 AG942 MGT001371 APH106 ECON1051 MA4AM MA3Z7 NUMBER THEORY REST0006 SESS0026 STAT0023 TABL 2741 CMP4008Y EFIM10015 Acct 560 PSYC735 GEOG 5 ACCT3104 CSEC5616 PHIL1012 ITLS6111 ECON8022 5CCM226A compX123 MA4XJ R001 EWB LLP120 COMP643 ENG 103 ECON 23950 ACCT 207 FINC 616 Bio99 Math 137A Psychology BCPM0054 MATH20212 EC3450 CENV6141 PSY 205 CE335 CS365 Calculus ACSD INFT6800 MCIT INFO6030 PHYS 111LV H63EEJ ECON2151 ENGI 2181 MTRX1701 PLIT08009 CS240 Math 270 CMP7000A CMP-7038B Math 263 LAW3016 CS371 STAT 334 FIN0008 ACC3004 STAT 371 ​COMP1921 MTH201 ACCT2000 UN3412 CIV6000 MATHS 3021 MATE7014 ECON1 CIBC6035 LAW121G COMM 321 CSC8204R TTM2033 CSCI 2600 MAE 115 LNG302 CENV6175W1 ECON30019 CPS2015 BST969 DDES9015 AMATH242 CS 1151 N1592 ACCT600 EG503G SMM519 EC318 Ecstatistics EEE 471 MARK2051 Topology COM00037H ACCT3000 MTHM501 EG501V COM00055H BUSN9085 GGR273 MATH 2B ITD102 MA214 ECON0060 ECE3114 ECMM447 Mechanics CCT361H5S BSAN3210 WELF2001 Fin3001 RESP5001 STA 137 Linguistics 101 CSCI368 CMPSC497 LUBS3655 OMBA 5355 BUSM2562 ​BFF5902 CYBR7001 CULT2005 ECON 7720 MTRX3760 FSE598 ​LUE1001 MGSC 7100 CS3240 COSC7500 ACCT3091 INFOGR TRAN 2080 DESIGN CIVE215001 CIVE2360 CPSC 8430 FNCE2003 Stoichiometry INFS6023 Quantitative Method Math 21D ECE5179 Design ​ECON8022 Chemical DMS 213 ADM1370 BFF5902 BFF2701 AGRI10051 AMN400 ECON7322 ​FINC5090 ​MATH1052 ​SciComp Elec4622 ​FINS5516 ENG 101 MMM143 CS431CA2 BSAN2201 EBSC7070 ES9v9-10 PLIN0009 DATA 8 MATH 101 BCPM0097 ​MAT 237 BISM2201 EEE8099 Physics 11 ECE2238B Biostat 234 ​ECON30009 INF10025 CHEM252 CCMA4008 CIVE 3007 STAT0031 ELEC ENG 3108 MGMT90140 CHEM 252 STAT3016 MMM095 ​MMGT6016 Legal STAT 311 ELEC9764 COMP 370 TCS 3053 EECS 1021 FIT1051 C/C++ EE3C COMP2003J COM398 COMP51915 Simulator FIN42330 8PRO102 PGD ELEC5160 XJCO1921 COMP5123M INFS 2042 CHE1148H ECA5313 IN3329 Microsoft Modeling B15 2TT COMP3687 COMP2432 DMS2030 KAN4TSF CFKG CSE 5322 CSE 445 STATS 3DA3 CHC5223 COMP1048 EL3105 EECS 3221 CS231 ME 4434 ITP4510 ITP4501 CSCI 5708 Controller EECE 5699 CPEN 231L CPN programs CSC8016 MATU9D2 ENGN4213 2. SVM. (10 points) Consider an SVM classifier using RBF kernel, and finish the following python code to search for the optimum kernel parameter γ. You are required to calculate the RBF function and search yourself, instead of using functions in Scikit-Learn (sklearn). def parameter search(gamma list ,Xtr,ytr,Xts,yts ,...): """ Parameters gamma list: A list of RBF kernel parameter gamma to search for Xtr: Training data, shape (ntr,nrow,ncol) ytr: Training labels, shape (ntr,) Xts: Test data, shape (nts,nrow,ncol) yts: Test labels, shape (nts,) ... Add other parameters as needed Returns: gamma optimum: The gamma with minimum error on Xts, yts """ ... return gamma optimum
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