Use of LAG - SQL Server 2014 - sql

Question

I need to calculate a column (Transparencia) as the sum of the previous row's Transparencia and Dif values. Initially, only the first row has a value in the Transparencia column:
Account ------ Year_ ---- Month_ ---- Transparencia ---- Dif
--------------------------------------------------------------
'4030003'------ 2018 ---- 5 ---- 100 ---- -2
'4040001'------ 2018 ---- 5 ---- null ---- -4
'4040002'------ 2018 ---- 5 ---- null ---- 3
...
Account(N-1)------ 2018 ---- 5 ---- x ---- 8
Account(N)------ 2018 ---- 5 ---- x + 8 ---- 11
The aim is to get the following:
Account ------ Year_ ---- Month_ ---- Transparencia ---- Dif
--------------------------------------------------------------
'4030003'------ 2018 ---- 5 ---- 100 ---- -2
'4040001'------ 2018 ---- 5 ---- 98 ---- -4
'4040002'------ 2018 ---- 5 ---- 94 ---- 3
...
Account(N-1)------ 2018 ---- 5 ---- x ---- 8
Account(N)------ 2018 ---- 5 ---- x + 8 ---- 11
Where:
98 = 100 + (-2) -> (Transparencia from previous row plus Dif from previous row)
94 = 98 + (-4) -> (Transparencia from previous row plus Dif from previous row)
x = 'Transparencia' from the previos row + 'Dif' from the previous row
x + 8 = 'Transparencia' from the previos row + 8 ('Dif' from the previous row)
The solution I tried was:
select
tmp.Account, tmp.Year_, tmp.Month_,Dif,
case
when Transparencia is null
then (lag(Transparencia, 1, 0) over (order by Account) -
lag(Dif, 1, 0) over (order by Account))
else Transparencia
end Transparencia
from
(select
Account,
nryear as Year_, nrperiod as Month_,
Dif, Transparencia
from
repaca
where
nrperiod = 5) tmp
However, this returns the following result:
Account ------ Year_ ---- Month_ ---- Transparencia ---- Dif
'4030003'------ 2018 ---- 5 ---- 100 ---- -2
'4040001'------ 2018 ---- 5 ---- 98 ---- -4
'4040002'------ 2018 ---- 5 ---- null ---- 3
I need to achieve this using just a SELECT, not a stored procedure or alike. Any help will be appreciated.
Thanks in advance

Answer 1

You don't want lag(). You want cumulative sum. Because the value is NULL, you can simplify getting the first value by using max(). So:
select r.*,
(max(Transparencia) over () +
sum(diff) over (order by year, month)
) as new_Transparencia
from repaca r;
You could also phrase this as:
select r.*,
sum(coalesce(Transparencia, 0) + diff) over (order by year, month) as new_Transparencia
from repaca r;
EDIT:
The above is using the wrong ordering. This seems to be:
select r.*,
(max(Transparencia) over (partition by year, month) +
sum(diff) over (partition by year, month order by account)
) as new_Transparencia
from repaca r;