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When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a Gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For waiting until one bin is filled, since all bins are empty the rate is $n$. The waiting time for a second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also Intuition about the coupon collector problem approaching a Gumbel distribution on Cross Validated.

This is of course not general.

If we use $X_i = N(\mu = 1/i, \sigma^2 = 1/i^2)$ then a (properly scaled) sum will approach a normal distribution.

Does anyone know if the Gumbel can occur as a limit distribution for such a sum?

When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a Gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For waiting until one bin is filled, since all bins are empty the rate is $n$. The waiting time for a second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also Intuition about the coupon collector problem approaching a Gumbel distribution on Cross Validated.

This is of course not general.

If we use $X_i = N(\mu = 1/i, \sigma^2 = 1/i^2)$ then a (properly scaled) sum will approach a normal distribution.

That is a trivial example but there are more cases that will converge to a normal distribution. The relevant condition that needs to be fulfilled is the Lyapunov condition.

When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a Gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For the first bin, since all bins are empty the rate is $n$. The waiting time for the second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also Intuition about the coupon collector problem approaching a Gumbel distribution on Cross Validated.

When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a Gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For waiting until one bin is filled, since all bins are empty the rate is $n$. The waiting time for a second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also Intuition about the coupon collector problem approaching a Gumbel distribution on Cross Validated.

This is of course not general.

If we use $X_i = N(\mu = 1/i, \sigma^2 = 1/i^2)$ then a (properly scaled) sum will approach a normal distribution.

When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For the first bin, since all bins are empty the rate is $n$. The waiting time for the second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also https://stats.stackexchange.com/questions/547372/intuiton-about-the-coupon-collector-problem-approaching-a-gumbel-distribution

When we have $n$ exponential distributed variables $X_i \sim Exp(\gamma = i)$, (with expectation $1/i$ and variance $1/i^2$) then the sum

$$S = \sum_{i=1}^n (X_i - 1/i)$$

approaches a Gumbel distribution.

There is a connection between this sum and the maximum order statistic.

We can see this sum as the waiting time for filling $n$ bins when the filling of the bins is a Poisson process.

• Approach with the sum. The waiting time between the filling of bins bin is exponential distributed. For the first bin, since all bins are empty the rate is $n$. The waiting time for the second bin to be filled is when $n-1$ bins are empty and the rate will be $n-1$, and so on...
• Approach with the maximum. We can consider the waiting times for filling each individual bin. The waiting time to fill all bins is equal to the maximum of the individual waiting times.

The distribution of the maximum of exponential distributed variables approaches a Gumbel distribution. Therefore the expression in terms of a sum, which has an equal distribution, will also approach the Gumbel distribution.

See also Intuition about the coupon collector problem approaching a Gumbel distribution on Cross Validated.