Return to Answer

Fiddle

Yet another way is to partition the cumulative sum, since the case expression only accounts for init <> 1, we will get the running sum for the rest. We then need to add the offset:

with classify (ts, value, init) as 
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
)
select ts, value
     , case when init = 1 
            then value
               -- cumulative sum for init = 0 
            else sum(value) over (partition by init order by ts)
               -- add offset
               + (select max(value) from classify where init = 1)
       end as valueA
from classify
order by ts  
;

Fiddle

Surely there must be a more elegant solution, but this shows that it is possible, so I'll leave the esthetics as an exercise;-)

EDIT: A slightly shorter version:

Fiddle

We don't really need the CTE:s for first_day, rem_days so a slight simplification is:

Which can be further simplified as:

with classify (ts, value, init) as 
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
)
select ts, value
     , case when init = 1 
            then value
            else sum(value) over (order by ts)
               -- remove all but last row from offset 
               - (select sum(value)-max(value) 
                  from classify where init = 1)
       end as valueA
from classify
order by ts  
;

Fiddle

Not the most elegant solution, but I believe we should divide it into two cases, first day or remaining days:

with classify (ts, value, init) as 
-- classify rows as first_day or remaining day
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
), first_day as ( 
  select ts, value, value as valueA
  from classify 
  where init = 1
), rem_days as (
  select ts, value, sum(value) over (order by ts) as valueA
  from classify
  where init = 0
)
select ts, value, valueA from first_day
union all
select ts, value, valueA + (select max(value) from first_day) 
from rem_days
order by ts
;

Surely there must be a more elegant solution, but this shows that it is possible, so I'll leave the esthetics as an exercise;-)

EDIT: A slightly shorter version:

with classify (ts, value, init) as 
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
)
select ts, value
     , case when init = 1 then value
                          else sum(value) over (order by ts) 
                          - (select sum(value) from classify where init = 1)
                          + (select max(value) from classify where init = 1)
       end as valueA
from classify
order by ts  

Fiddle

Not the most elegant solution, but I believe we should divide it into two cases, first day or remaining days:

with classify (ts, value, init) as 
-- classify rows as first_day or remaining day
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
), first_day as ( 
  select ts, value, value as valueA
  from classify 
  where init = 1
), rem_days as (
  select ts, value, sum(value) over (order by ts) as valueA
  from classify
  where init = 0
)
select ts, value, valueA from first_day
union all
select ts, value, valueA + (select max(value) from first_day) 
from rem_days
order by ts
;

Surely there must be a more elegant solution, but this shows that it is possible, so I'll leave the esthetics as an exercise;-)

EDIT: A slightly shorter version:

with classify (ts, value, init) as 
(
  select ts, value
       , first_value(date(ts)) over (order by ts) = date(ts) as init
  from MyTable
)
select ts, value
     , case when init = 1 
       then value
       else sum(value) over (order by ts) 
            - (select sum(value) from classify where init = 1)
            + (select max(value) from classify where init = 1)
       end as valueA
from classify
order by ts  

Fiddle