2017-09-13

Sep 13 In-Class Exercise.

Post to this thread your solutions to the in-class exercise.
Best, Chris
Post to this thread your solutions to the in-class exercise. Best, Chris

-- Sep 13 In-Class Exercise
delta = 1/1000 epsilon = 1/(n + 1)
(Edited: 2017-09-13)
'''delta''' = 1/1000 '''epsilon''' = 1/(n + 1)

-- Sep 13 In-Class Exercise
By the proof, the number of examples needed is:
.
`m=max(\frac{144ln\delta}{2},m_i)`
.
If `\delta=\frac{1}{1000}`, then
.
`m=max(\frac{144ln(\frac{2}{1000})}{2},m_1)`
.
The number of examples needed is:
.
`2\frac{m}{\epsilon}`
.
Given `\epsilon=\frac{1}{n+1}`, this means the number of examples is:
.
`2max(144ln(\frac{2}{1000})}{2}(n+1),m_1)`
.
It was given that:
.
`m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}`
(Edited: 2017-09-13)
By the proof, the number of examples needed is: . @BT@m=max(\frac{144ln\delta}{2},m_i)@BT@ . If @BT@\delta=\frac{1}{1000}@BT@, then . @BT@m=max(\frac{144ln(\frac{2}{1000})}{2},m_1)@BT@ . The number of examples needed is: . @BT@2\frac{m}{\epsilon}@BT@ . Given @BT@\epsilon=\frac{1}{n+1}@BT@, this means the number of examples is: . @BT@2max(144ln(\frac{2}{1000})}{2}(n+1),m_1)@BT@ . It was given that: . @BT@m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}@BT@

-- Sep 13 In-Class Exercise
probability is 1 - 1/1000 = 1 - delta
delta = 1/1000, epsilon = 1/(n+1)
number of examples =2m/epsilon = 2m (n+1)
m= max(144ln (1/1000) / 2, m1)
m1 = ...
(Edited: 2017-09-13)
probability is 1 - 1/1000 = 1 - delta delta = 1/1000, epsilon = 1/(n+1) number of examples =2m/epsilon = 2m (n+1) m= max(144ln (1/1000) / 2, m1) m1 = ...

-- Sep 13 In-Class Exercise
delta = 1-(1/1000) e = 1/(n+1) m = max((144*lnδ)/2,m1) m1 = 144(n+1)2/25(kε)2
delta = 1-(1/1000) e = 1/(n+1) m = max((144*lnδ)/2,m1) m1 = 144(n+1)2/25(kε)2

-- Sep 13 In-Class Exercise
m=max(144 *lv (2/1000)/2,m1) m1=144 (n+1)^4/(25/36) number of samples = 2*m/epsilone
(Edited: 2017-09-13)
m=max(144 *lv (2/1000)/2,m1) m1=144 (n+1)^4/(25/36) number of samples = 2*m/epsilone

-- Sep 13 In-Class Exercise
`m=max(\frac{144ln\delta}{2},m_i)`
`m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}`
Given,
`\delta=\frac{1}{1000}`
`\epsilon=\frac{1}{n+1}`
Thus,
`m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)`
`m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }`
Since, ln(1/1000) / 2 < 0,
`m = (144*(n+1)^4) / 25*k^2`
Thus, no. of samples
`= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)`
`= (288/ 25) * ( (n+1)^5 / k^2)`
k = 1 / (c * 2) = 1 / (3 * 2) = 1 / 6
Thus, no. of samples:
`= (288 * 36/ 25) * ( (n+1)^5 )`
 
(Edited: 2017-10-02)
@BT@m=max(\frac{144ln\delta}{2},m_i)@BT@ @BT@m_1=\frac{144(n+1)^2}{25(k\epsilon)^2}@BT@ Given, @BT@\delta=\frac{1}{1000}@BT@ @BT@\epsilon=\frac{1}{n+1}@BT@ Thus, @BT@m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)@BT@ @BT@m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }@BT@ Since, ln(1/1000) / 2 < 0, @BT@m = (144*(n+1)^4) / 25*k^2@BT@ Thus, no. of samples @BT@= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)@BT@ @BT@= (288/ 25) * ( (n+1)^5 / k^2)@BT@ k = 1 / (c * 2) = 1 / (3 * 2) = 1 / 6 Thus, no. of samples: @BT@= (288 * 36/ 25) * ( (n+1)^5 )@BT@

-- Sep 13 In-Class Exercise
No of examples to be looked at= 2`\frac{m}{e}`——(1)
Given:
 delta=1/1000
e=1/n+1
Now, we know that
m = max(144ln delta/2,m1) Since first term would be negative,
m=m1
Now, m1=(144(n+1)^2)/(25(ke)^2)
Substituting in eq 1, we have
No of examples=(288(n+1)^3)/(25(ke)^2)
(Edited: 2017-09-13)
No of examples to be looked at= 2@BT@\frac{m}{e}@BT@——(1) Given:
 delta=1/1000 e=1/n+1 Now, we know that m = max(144ln delta/2,m1) Since first term would be negative, m=m1 Now, m1=(144(n+1)^2)/(25(ke)^2) Substituting in eq 1, we have No of examples=(288(n+1)^3)/(25(ke)^2)

-- Sep 13 In-Class Exercise
`δ = 1/1000`
`ε = 1 / (n + 1)`
`m1 =144*(n+1)^2/(25*(kε)^2)`
Substituting value of ε:
`m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)`
`m1 =(144/25) *( (n+1)^4 / k^2)`
`m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }`
As ln(1/1000) / 2) < 0 and the second term is always positive Therefore `m = (144*(n+1)^4) / 25*k^2`
No of examples to be used = `2 * (m/ ε)`
`= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)`
`= (288/ 25) * ( (n+1)^5 / k^2)`
Calculating value of k -> `c = 1 / 2*k`
c = 3 -> k = 1 / 6
Hence no of examples to be used `= (288/ 25) * ( (n+1)^5 / (1/6)^2)`
`= (288 * 36/ 25) * ( (n+1)^5 )`
(Edited: 2017-09-13)
@BT@δ = 1/1000@BT@ @BT@ε = 1 / (n + 1)@BT@ @BT@m1 =144*(n+1)^2/(25*(kε)^2)@BT@ Substituting value of ε: @BT@m1 =144*(n+1)^2 / (25*(k * (1/(n+1))^2)@BT@ @BT@m1 =(144/25) *( (n+1)^4 / k^2)@BT@ @BT@m = max{ (144 * ln(1/1000) / 2), ((144*(n+1)^4) / (25*k^2)) }@BT@ As ln(1/1000) / 2) < 0 and the second term is always positive Therefore @BT@m = (144*(n+1)^4) / 25*k^2@BT@ No of examples to be used = @BT@2 * (m/ ε)@BT@ @BT@= 2 * (144*(n+1)^4 / 25*k^2) * (n + 1)@BT@ @BT@= (288/ 25) * ( (n+1)^5 / k^2)@BT@ Calculating value of k -> @BT@c = 1 / 2*k@BT@ c = 3 -> k = 1 / 6 Hence no of examples to be used @BT@= (288/ 25) * ( (n+1)^5 / (1/6)^2)@BT@ @BT@= (288 * 36/ 25) * ( (n+1)^5 )@BT@

-- Sep 13 In-Class Exercise
number of examples= 2m /epsilon
m = max(144lnδ/2,mi)
m1 = 144(n+1)4 / 25 k2
so, number of examples = 288 (n+1)5 / 25(1/6)2
number of examples= 2m /epsilon m = max(144lnδ/2,mi) m1 = 144(n+1)4 / 25 k2 so, number of examples = 288 (n+1)5 / 25(1/6)2
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