how to take the natural log of a product ($prod$)
I have this likelihood function, $L(theta) = prod _{i=1}^nleft(frac{1}{theta :}e^{-frac{x_i}{theta :}}right)$. I'm trying to take the natural log, $ln(L(theta))$, but I'm not sure how this works with respect to $prod$. Does anyone know what the process for this log is?
logarithms products
edited Jan 6 at 14:03
amWhy
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
add a comment |
I have this likelihood function, $L(theta) = prod _{i=1}^nleft(frac{1}{theta :}e^{-frac{x_i}{theta :}}right)$. I'm trying to take the natural log, $ln(L(theta))$, but I'm not sure how this works with respect to $prod$. Does anyone know what the process for this log is?
logarithms products
edited Jan 6 at 14:03
amWhy
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
2
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34
add a comment |
I have this likelihood function, $L(theta) = prod _{i=1}^nleft(frac{1}{theta :}e^{-frac{x_i}{theta :}}right)$. I'm trying to take the natural log, $ln(L(theta))$, but I'm not sure how this works with respect to $prod$. Does anyone know what the process for this log is?
logarithms products
edited Jan 6 at 14:03
amWhy
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
I have this likelihood function, $L(theta) = prod _{i=1}^nleft(frac{1}{theta :}e^{-frac{x_i}{theta :}}right)$. I'm trying to take the natural log, $ln(L(theta))$, but I'm not sure how this works with respect to $prod$. Does anyone know what the process for this log is?
logarithms products
logarithms products
edited Jan 6 at 14:03
amWhy
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
edited Jan 6 at 14:03
amWhy
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
edited Jan 6 at 14:03
amWhy
192k28225439
edited Jan 6 at 14:03
amWhy
192k28225439
edited Jan 6 at 14:03
amWhy
192k28225439
192k28225439
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
asked Nov 4 '17 at 22:31
lmotl3lmotl3
32
32
2
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34
add a comment |
2
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34
2
2
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34
add a comment |
3 Answers
3
active
oldest
votes
The log of a product is the sum of logs of the things inside the product.
So
$$ln L(theta)=sum_{i=1}^n lnleft(frac{1}{theta}e^{-x_i/theta}right)=sum_{i=1}^n left(lnleft(frac{1}{theta}right)-frac{x_i}{theta}right)$$
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
add a comment |
$$ln{(L(theta))} =sum_{i=1} ^n lnleft(frac{1}{theta}e^{-frac{x_i}{theta}}right) $$
This is because of the product rule for logarithms, that says that $log_a (BC) = log_a (B) + log_a (C)$.
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
add a comment |
The result is trivial
$$-nln(theta) - frac{1}{theta}sum_{i=1}^{n}x_i$$
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
The log of a product is the sum of logs of the things inside the product.
So
$$ln L(theta)=sum_{i=1}^n lnleft(frac{1}{theta}e^{-x_i/theta}right)=sum_{i=1}^n left(lnleft(frac{1}{theta}right)-frac{x_i}{theta}right)$$
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
add a comment |
The log of a product is the sum of logs of the things inside the product.
So
$$ln L(theta)=sum_{i=1}^n lnleft(frac{1}{theta}e^{-x_i/theta}right)=sum_{i=1}^n left(lnleft(frac{1}{theta}right)-frac{x_i}{theta}right)$$
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
add a comment |
The log of a product is the sum of logs of the things inside the product.
So
$$ln L(theta)=sum_{i=1}^n lnleft(frac{1}{theta}e^{-x_i/theta}right)=sum_{i=1}^n left(lnleft(frac{1}{theta}right)-frac{x_i}{theta}right)$$
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
The log of a product is the sum of logs of the things inside the product.
So
$$ln L(theta)=sum_{i=1}^n lnleft(frac{1}{theta}e^{-x_i/theta}right)=sum_{i=1}^n left(lnleft(frac{1}{theta}right)-frac{x_i}{theta}right)$$
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
answered Nov 4 '17 at 22:36
A. GoodierA. Goodier
3,57151326
3,57151326
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
add a comment |
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
I feel like you owe @Wojowu some credit for this. The first clause mirrors the comment exactly.
– Chase Ryan Taylor
Nov 4 '17 at 22:41
add a comment |
$$ln{(L(theta))} =sum_{i=1} ^n lnleft(frac{1}{theta}e^{-frac{x_i}{theta}}right) $$
This is because of the product rule for logarithms, that says that $log_a (BC) = log_a (B) + log_a (C)$.
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
add a comment |
$$ln{(L(theta))} =sum_{i=1} ^n lnleft(frac{1}{theta}e^{-frac{x_i}{theta}}right) $$
This is because of the product rule for logarithms, that says that $log_a (BC) = log_a (B) + log_a (C)$.
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
add a comment |
$$ln{(L(theta))} =sum_{i=1} ^n lnleft(frac{1}{theta}e^{-frac{x_i}{theta}}right) $$
This is because of the product rule for logarithms, that says that $log_a (BC) = log_a (B) + log_a (C)$.
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
$$ln{(L(theta))} =sum_{i=1} ^n lnleft(frac{1}{theta}e^{-frac{x_i}{theta}}right) $$
This is because of the product rule for logarithms, that says that $log_a (BC) = log_a (B) + log_a (C)$.
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
answered Nov 4 '17 at 22:37
WaveXWaveX
2,5002721
2,5002721
add a comment |
add a comment |
The result is trivial
$$-nln(theta) - frac{1}{theta}sum_{i=1}^{n}x_i$$
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
add a comment |
The result is trivial
$$-nln(theta) - frac{1}{theta}sum_{i=1}^{n}x_i$$
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
add a comment |
The result is trivial
$$-nln(theta) - frac{1}{theta}sum_{i=1}^{n}x_i$$
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
The result is trivial
$$-nln(theta) - frac{1}{theta}sum_{i=1}^{n}x_i$$
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
answered Nov 4 '17 at 22:40
Gevorg HmayakyanGevorg Hmayakyan
605317
605317
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
add a comment |
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
This doesn't look correct. How did you reach this answer?
– MBorg
Sep 15 '18 at 14:44
add a comment |
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2
Log of a product is a sum of logs.
– Wojowu
Nov 4 '17 at 22:34