Is the exponential distribution a good model for this data?

Question

I'm trying to determine if the exponential distribution is a good model for a data set that I'm exploring. It doesn't have to be precise. I'm using the data for capacity planning (if it's a good fit) and for my own learning. I found a data set that I have ready access to and I wanted to test to see it fit a known distribution.

To test if it is a good fit, I've generated 2 plots from a data set with about 1.5M entries.

In the first plot, I take the data and generate an empirical CDF and plot the log of the CCDF and the linear model using geom_smooth. I'm doing this because if it is exponential (y ~= e^−λx), then the log of both sides gives log y ~= -λx which should be a straight line with slope -λ.

data <- sort(data) 
valid <- data[1:(length(data)*0.96)]  
ecdf <- (1:length(valid))/length(valid)
df <- data.frame(cbind(x=valid, y=ecdf))  

# if this is exponential, get the slope and the x intercept
lm(log(1-y)~x, data=df, subset=seq(1,length(valid)-1))

ggplot(aes(x=x, y=log(1-y)), data=df[1:length(valid)-1,]) +
    geom_line(color="blue") +
    geom_smooth(method="lm", color="red", size=0.3) +
    labs(x="Data", y="CCDF of Data Log(y)", title="Line of Best Fit on Log(CCDF) of Data")

Here is the result. It mostly fits except at the tail. Should I conclude that the exponential distribution is a (mostly) good fit?

Edit: here is the summary data from the model generated by lm for the Log(CCDF) plot:

Residuals:
    Min      1Q  Median      3Q     Max 
-9.7550 -0.0750  0.0044  0.1171  0.1328 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept)  1.239e-01  2.388e-04   518.8   <2e-16 ***
x           -1.868e-03  2.989e-07 -6248.6   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 0.1907 on 1473365 degrees of freedom
Multiple R-squared: 0.9636,     Adjusted R-squared: 0.9636 
F-statistic: 3.904e+07 on 1 and 1473365 DF,  p-value: < 2.2e-16 

Next I attempted to generate a Q-Q plot by generating a data set that is the same size as the empirical data set (about 1.5M) using a function that generates random numbers from an exponential distribution. From the lm above I used the x output and switched the sign as my lambda. I sorted both of them and plotted the sampled data as x and the empirical data as y.

Here is the result. Again, it's not quite x=y. Should I conclude that the exponential distribution is a good enough fit?

Edit: here is the summary data for the model from the Q-Q plot as generated by lm.

Residuals:
    Min      1Q  Median      3Q     Max 
-49.256  -9.214  -0.069  10.060 268.769 

Coefficients:
              Estimate Std. Error t value Pr(>|t|)    
(Intercept) -1.757e+01  4.086e-02  -429.9   <2e-16 ***
x            1.264e+00  6.078e-05 20803.7   <2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 

Residual standard error: 32.82 on 1488715 degrees of freedom
Multiple R-squared: 0.9966,     Adjusted R-squared: 0.9966 
F-statistic: 4.328e+08 on 1 and 1488715 DF,  p-value: < 2.2e-16 

Thanks everyone for all of the helpful comments!

Answer 1

It depends on what you want to use this for. I can easily imagine situations in capacity planning where you would be most interested in extreme occurrences, as these peak events are what strains capacity most. If that is the case, then your tail behaviour would be a problem. I can also imagine other situations where the system is somewhat flexible so that they can absorb short peaks, in which case you would be more interested in typical behaviour. In that case your model may work.

Answer 2

I know this is not really a direct answer to your question, but I was looking for a qq-plot function for the exponential distribution in R and found this. It may be helpful for other people browsing this page with the goal to find a qq-plot function for exponential distribution.

"qqexp" <-  function(y, line=FALSE, ...) { 
    y <- y[!is.na(y)]
    n <- length(y)
    x <- qexp(c(1:n)/(n+1))
    m <- mean(y)
    if (any(range(y)<0)) stop("Data contain negative values")
    ylim <- c(0,max(y))
    qqplot(x, y, xlab="Exponential plotting position",ylim=ylim,ylab="Ordered sample", ...)
    if (line) abline(0,m,lty=2)
    invisible()
  }