## How to use lognrnd?

Yaser Khojah

### Yaser Khojah (view profile)

さんによって質問されました 2019 年 3 月 11 日

### Torsten (view profile)

さんによって 編集されました 2019 年 3 月 12 日
Torsten

### Torsten (view profile)

さんの 回答が採用されました
I'm trying to create a random log-normal distributions for a vector . I have already read the lognrnd document as https://uk.mathworks.com/help/stats/lognrnd.html.and other people who ask a similar question but I'm a bit confused since the document is a bit confusing for me. So, I understand to use lognrnd as R = lognrnd(mu,sigma), both mu and sigma has to be mean and standard deviation of the normal distribution, let us call it Y. The corresponding lognormal distribution of X is Y. To use lognrnd from its given X (lognormal distribution) mean (m) and X variance (v), we should find the mu and sigma first by using these formula as
mu = log((m^2)/sqrt(v+m^2));
sigma = sqrt(log(v/(m^2)+1));
Now, this is clear for me but the confusing part in the document is the following sentence "If X is distributed lognormally with parameters µ and σ, then log(X) is distributed normally with mean µ and standard deviation σ"
Should not be if X is distributed lognormally with parameters m and v since µ and σ represent only the normal distribution?
I have listed what I'm trying to do just to be clear about my understanding and for confirmation about how to estimate random distribution from given mean and standard deviation of the log normal distribution:
m = 14; % mean of the lognormal
std_d = 7; % standard deviaiton of the lognormal
v = std_d^2; % varaiance of the log
mu = log((m.^2)./ sqrt(v+m.^2)); % mu is the normal mean
sigma = sqrt(log(v./(m.^2)+1)); % sigma is the normal standard deviation
R = lognrnd(mu,sigma,10^6,1);
Thanks for your help and reading this.

#### 0 件のコメント

サインイン to comment.

## 1 件の回答

2019 年 3 月 11 日

### Torsten (view profile)

2019 年 3 月 11 日
採用された回答

It is common to say that X is lognormally distributed with parameters mu and sigma if it has the pdf
f(x) = 1/(x*sigma*sqrt(2*pi))*exp(-(log(x)-mu)^2/(2*sigma^2))
If you now calculate the pdf of log(X), you'll see that it equals
g(y) = 1/(sigma*sqrt(2*pi))*exp(-(y-mu)^2/(2*sigma^2))
Thus log(X) is normally distributed with parameters mu and sigma.
To calculate 10^6 random numbers from a lognormally distributed random variable with mean 14 and standard deviation 7, your code from above is correct.

Torsten

### Torsten (view profile)

2019 年 3 月 11 日
See my answer given. It is the "empirical proof" that the statement is correct.
Yaser Khojah

### Yaser Khojah (view profile)

2019 年 3 月 11 日
I just do not know what you mean by x and X?
Torsten

### Torsten (view profile)

2019 年 3 月 12 日
If X is a continuous real-valued random variable and x is a value in the real numbers, then the cumumlative density function F of the random variable X at x is defined as
F(x) = P(X<=x)
where P(X<=x) is the probability that the random variable X produces a value y that is smaller than or equal to x.
I think you will have to learn the basics of probability theory first before playing around with complicated distributions.

サインイン to comment.