How to calculate a value based on the previous row's value - sql

Question

Having the following fields in a table...
+---------+---+---+
| myTime | x | y |
+---------+---+---+
| 13:00 | 0 | 0 |
| 13:05 | 2 | 1 |
| 13:10 | 4 | 2 |
| 13:15 | 1 | 3 |
+---------+---+---+
I need to generate a third one (z) as follows...
+---------+---+---+---+
| myTime | x | y | z |
+---------+---+---+---+
| 13:00 | 0 | 0 | 0 |
| 13:05 | 2 | 1 | 1 |
| 13:10 | 4 | 2 | 3 |
| 13:15 | 1 | 3 | 1 |
+---------+---+---+---+
In the first row z will have a value of 0 and in the next ones, z will be calculated as x-y + (previous row's) z.
I've tried using the row number for each record and LAG to try reading values from previous rows...
WITH rows_sorted AS
(SELECT *, ROW_NUMBER() OVER (ORDER BY myTime) AS row_num
FROM table)
SELECT myTime, x, y
IF(row_num = 1, 0, x - y + LAG(z, 1) OVER (ORDER BY row_num)) AS z
FROM rows_sorted
ORDER BY row_num
...but evidently wouldn't work as in LAG(z, 1), z has not been generated yet.
Any suggestion on how such a thing can be done? I'm using standard SQL in Google BigQuery
Thanks in advance
Since the text above oversimplifies the real calculation, here's a closer approach to what I need to achieve:
+---------+----+----+----+
| myTime | x | y | z |
+---------+----+----+----+
| 13:00 | 15 | 22 | 0 |
| 13:05 | 7 | 21 | 0 |
| 13:10 | 7 | 5 | 2 |
| 13:15 | 9 | 16 | 0 |
| 13:20 | 14 | 5 | 9 |
+---------+----+----+----+
Where z for each row is calculated as follows:
WHEN row_number() = 1 THEN z = 0 (already achieved thanks to the
answer below)
WHEN x+(previous row's)z < y THEN z = 0
WHEN x+(previous row's)z >= y THEN z = x+(previous row's)z - y

Answer 1

Hmmm . . . You can get what you want using:
select t.*,
sum(x - y) over (order by mytime) as z
from t;
The first row has values of 0 for all the columns, so this works for your sample data. If you wanted to explicitly set it to 0, then:
select t.*,
(case when row_number() over order by mytime) = 1
then 0
else sum(x - y) over (order by mytime) - first_value(x - y) over (order by mytime)
end) as z
from t;
This subtracts out the value from the first row from the cumulative sum. However, that seems unnecessary.