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The given problem is as follows:

Recall that $\log 2 = \int_0^1 1/(x+1) dx$. Hence, by using a uniform $(0,1)$ generator, approximate $\log 2$. Obtain an error of estimation in terms of a large sample 95% confidence interval. If you have access to the statistical package R, write an R function for the estimate and the error of estimation. Obtain your estimate for 10,000 simulations and compare it to the true value.

My answer:

$$\int_0^1 1/(x+1) dx = (1-0)\int_0^1 1/(x+1) dx/(1-0) = \int_0^1 1/(x+1) f(x) dx = E(1/(x+1))$$

Where f(x)=1, 0

And then I calculated log 2 from the calculator and got 0.6931471806

From R, I got 0.6920717

So, from the weak law of large numbers, we can see that the sample mean is approaching the actual mean as n gets larger.

My Question:

Is my answer correct? Can I use the calculator to approximate log 2?

Thanks in advance

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1  
You are supposed to use $\log 2 = \int_0 ^1 \frac {1}{x+1}\, dx$ and the uniform distribution, to approximate $\log 2$. Not supposed to type $\log 2$ into your calculator and see what you get... So your answer is wrong (without knowing what the numerical value of $\log 2$ is.) –  Calvin Lin Jan 29 '13 at 8:50
    
@CalvinLin I multiplied by the integral with (1-0) because I divided dx by (1-0) (since I want to turn the integral into an expectation...so I need to multiply it by some function, and I chose that function to be f(x)=1, 0<x<1. The problem is...that when I calculate the expectation, I get log 2...so it doesn't really help much unless I use the calculator. Can you give me a hint if my answer is wrong? –  user58289 Jan 29 '13 at 9:06
    
I don't see an answer of yours to the problem, so I can't comment if it's right or wrong. I believe that you are supposed to generate several random variables, and then calculate the mean of $\frac {1}{x+1}$ to approximate $\log 2$. (I believe that) You can use the weak law of large numbers to tell you how many variables you need, in order to get a 95% CI. –  Calvin Lin Jan 29 '13 at 9:16
    
@CalvinLin Actually, I'm sorry, I forgot to tell you...I know I didn't compute the confidence interval yet. I was just wondering if I was correct so far? –  user58289 Jan 29 '13 at 9:23
    
I edited your question to make the distinction clearer between the given problem and your own work, as I understood it. Please check that I did not change the intent. –  Rahul Jan 29 '13 at 10:32

1 Answer 1

up vote 1 down vote accepted

You are supposed to generate $n = 10000$ random variables in R. You could obtain them by typing

x<-runif(10000)

This generate $x_1, \ldots, x_n$ from a uniform $U \left( 0, 1 \right)$.

To get an estimate of $\log \left( 2 \right)$, you compute the sum $$ \frac{1}{n}\sum_{i = 1}^n \frac{1}{x_i + 1} $$ which converges by the law of large number to $E \left[ \frac{1}{1 + x} \right] = \log \left( 2 \right)$

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Yes I did get the sample mean for a size of 10,000 from R. It was 0.6920717. I then compared it with what I got from the calculator. Is that correct? –  user58289 Jan 29 '13 at 10:49
    
Yes. However, don't forget that if you run the R code again, you will obtain a different answer than 0.6920717 (because the sample is random), but still an answer close to $\log(2)$. That is why you will have to look at the confidence interval to assess how accurate the estimator is. –  Learner Jan 29 '13 at 10:58
    
Ok thanks. Actually, I was just asking if what I did so far was correct...I know I had to find the confidence interval too, but that wasn't my question. Anyways thanks. –  user58289 Jan 29 '13 at 11:19

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