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Why does plot(ecdf(1:1000)) produce a straight line? plot(ecdf(1:1000)) in R

Since Fn($x_n$) = $x_1$/(total sum) +$x_2$/(total sum) +...+$x_n$/total sum = ($x_1+x_2+x_3+...+x_n$)/total sum. the fact that Fn(200) roughly being equal to 0.2 and sum(0:200) roughly being equal to 0.4 seems to indicate that sum(1:200) is roughly half of sum(1:400), which isn't true, the two expressions being 20,000 and 80,000 respectively.

What am i misunderstanding?

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    $\begingroup$ Give a good look. It is a step function. $\endgroup$ – Khashaa Jul 4 '16 at 23:58
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The cumulative distribution function of a random variable $X$ has nothing to do with summing the random variable. It is

the probability that $X$ will take a value less than or equal to $x$.

And of course, the probability that a value randomly sampled from your vector $(1, \dots, 1000)$ is less than or equal to 200 is exactly half the probability that it is less than or equal to 400.

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Empirical cumulative distribution function is a cumulative sum of frequencies of observed $x_i$'s divided by total sample size. Your data is a vector of values from $1$ to $1000$, where each of the values appears exactly once. This means that your "variable" follows a discrete uniform distribution, that has a flat CDF.

As you can see on the example below, it'd be different if you used other imput data.

set.seed(123)

x <- sample(0:1000, 1e5, replace = TRUE)
y <- rnorm(1e5)

def <- par(mfrow = c(1,2))
plot(ecdf(x))
plot(ecdf(y))
par(def)

enter image description here

or

z <- c(1,2,5,7,12,14,19,25,100,250,300,301,500,800,900,901,1000)
plot(ecdf(z))

enter image description here

Notice that in the second example distances between different values are different so no matter that each value appeared only once, the line is curved.

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You can think about it mechanically, too.

The ECDF $\hat F$ evaluated at $x$ is the proportion of observations with value $x$ or below. Since you have exactly 1,000 observations $\{y_i\}_{i=i}^{1000}$, the difference between $\hat F(y_i)$ and $\hat F(y_{i+1})$ is always 0.001 for any $1 \le i < 1000$.

Moreover, your sample values are evenly spaced, so the difference between $y_i$ and $y_{i+1}$ is always 1. Therefore, for any $1 \le i < 1000$, the slope between $\left(y_i, \hat F(y_i)\right)$ and $\left(y_{i+1}, \hat F(y_{i+1})\right)$ is always $\frac{0.001}{1}$. A curve with a constant slope is just a straight line.

As for what you're misunderstanding, the $Fn$ you defined is definitely not the right formula. The denominator should be the number of observations, and the numerator should be the number of observations with value at or below $x_n$.

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The empirical distribution function of a sample $Y_1, ..., Y_n$ is defined as

$$ \widehat{F}(x) = \frac{1}{n} \sum_{i=1}^{n} \mathcal{I} \{ Y_i \leq x \} $$

In your data set, $Y_i = i$. So, $ \widehat{F}(x) = x/n$, for $x = 1, 2, ..., 1000$. Plotted the way you did, this looks like a linear function of $x$.

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    $\begingroup$ The final equation is incorrect. $\hat F$ is a step function, not a linear function. $\endgroup$ – whuber Jul 5 '16 at 14:06
  • $\begingroup$ @whuber, I never did thank you for that astute, and very substantively important, correction to my answer. Keep up the good work. $\endgroup$ – gammer Jan 9 '17 at 0:17

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