<p>The "proof" that you're looking for consists of a few steps.</p>

<ol>
<li><p>The second utility function is a log transformation of the first;</p></li>
<li><p>Log transformations are order preserving (monotone);</p></li>
<li><p>This #2 means that the each bundle is ranked the same from most preferred to least preferred;</p></li>
<li><p>Since preferences are ordinal and utility functions are representations of ordinal preferences these two utility functions are the same because they represent the same set of preferences. </p></li>
</ol>

<p>Hope that this helps.</p>

CC BY-SA 3.0
2017-04-28T16:11:12.780Z
(btw, it is not necessary to address the answers to the OP. This is evident) Also, there seems to be a convention in SE of not replying to off-topic questions. The idea is the the OP reformulates the question to make it on-topic.
luchonacho
CC BY-SA 3.0
2017-04-28T16:48:18.363Z
Lunchonacho, thanks. Noted it! It seemed pretty on topic to me and concrete. Granted it's not very politely formulated and OP comes off as lazy by not showing his or her work, but sometimes some of us are really stuck. Particularly, if they've never had to do something like this before.
Toby
CC BY-SA 3.0
2017-04-28T16:55:16.633Z
I agree that the topic itself is entirely on-topic. The problem is the person is not explicit on her/his own attempt. We do not want to encourage lazyness, but rather genuine effort. Sometimes a simple "I tried this but I'm stuck" is enough.
luchonacho
CC BY-SA 3.0
2017-04-28T19:34:24.287Z
I think more than monotonicity is at play here: logs are increasing functions. I.e. the worst outcome is not the best.
ahorn
CC BY-SA 3.0
2017-04-28T15:07:50.100Z
2017-04-28T16:11:21.987Z
2017-04-28T16:11:21.987Z
Toby
<p>How do you show that the utility function </p>

<p>$$U = (X + A)^p (Y + B)^q$$</p>

<p>gives the same result than</p>

<p>$$ U= p \log (X+A) + q \log (Y+B) $$</p>

CC BY-SA 3.0
2017-04-28T13:06:14.040Z
What do you mean by result? The result of the maximization? Btw the second utility function is the first one after applying logarithm and since it is a linear transformation the result of the maximization would be the same.
PostDocing
CC BY-SA 3.0
2017-04-28T15:02:24.790Z
You mean monotone transformation?
Toby
<utility>
Show result is the same with two utility functions
E G A
# Show result is the same with two utility functions

(Asked by: username_0 on 2017-04-28)

How do you show that the utility function

$$U = (X + A)^p (Y + B)^q$$

gives the same result than

$$ U= p \log (X+A) + q \log (Y+B) $$

**Tags:** utility
> **Comments:**
> **username_1**: What do you mean by result? The result of the maximization? Btw the second utility function is the first one after applying logarithm and since it is a linear transformation the result of the maximization would be the same.  
> **username_2**: You mean monotone transformation?  

---

## 1 answers

### Answer by username_2 on 2017-04-28. Score: 2

The "proof" that you're looking for consists of a few steps.

1. The second utility function is a log transformation of the first;
2. Log transformations are order preserving (monotone);
3. This #2 means that the each bundle is ranked the same from most preferred to least preferred;
4. Since preferences are ordinal and utility functions are representations of ordinal preferences these two utility functions are the same because they represent the same set of preferences.

Hope that this helps.

> **Comments:**
> **username_3**: (btw, it is not necessary to address the answers to the OP. This is evident) Also, there seems to be a convention in SE of not replying to off-topic questions. The idea is the the OP reformulates the question to make it on-topic.  
> **username_2**: Lunchonacho, thanks. Noted it! It seemed pretty on topic to me and concrete. Granted it's not very politely formulated and OP comes off as lazy by not showing his or her work, but sometimes some of us are really stuck. Particularly, if they've never had to do something like this before.  
> **username_3**: I agree that the topic itself is entirely on-topic. The problem is the person is not explicit on her/his own attempt. We do not want to encourage lazyness, but rather genuine effort. Sometimes a simple "I tried this but I'm stuck" is enough.  
> **username_4**: I think more than monotonicity is at play here: logs are increasing functions. I.e. the worst outcome is not the best.  

---

