observable which is made based on another hot observable is definitely hot? - rxjs

Question

Is this true? I'm currently a little confused:
observable which is made(merge, scan, map, filter... ) based on another hot observable is definitely hot?
Whatif I merged hot and cold observable together? is the new observable hot or cold?
Edit: test, result not expected
var source = Rx.Observable.fromEvent(window, 'mousemove')
.merge(Rx.Observable.from([1,2,3,4,5]))
.scan(acc=>++acc, 0);
setTimeout(()=>{
source.subscribe(x=>{
console.log('sub1', x)
})
}, 4000)
setTimeout(()=>{
source.subscribe(x=>{
console.log('------sub2', x)
})
}, 6000)
As paulpdaniels mentioned in his answer, mousemove should be a hot observable, so it should emit values even if there's no subscription.
But in the code example above, I delayed the subscription for 4 and 6 seconds. So when the application starts, I move my mouse around. When it reaches the 4th second, it logs out:
"sub1" 1
"sub1" 2
"sub1" 3
"sub1" 4
"sub1" 5
why it starts from 1?? doesn't it should have been accumulating values for 4 seconds? it should start from a bigger number from my understanding.
When it reaches the 6th second, it logs out:
"------sub2" 1
"------sub2" 2
"------sub2" 3
"------sub2" 4
"------sub2" 5
Why? doesn't it mean mousemove is cold observable?? or the problem is on scan operator??
PS: I'm using rxjs 5

Answer 1

That is correct. Hot simply means that it can produce events in the absence of subscribers. So if a source is hot then regardless of what comes after it, that source will remain hot.
Consider a simple example of a mouse event handler.
var mousemove = Rx.Observable.fromEvent(window, 'mousemove');
No matter what operators you attach to it, the source (mouse event handler) will produce a new event for every mouse movement. If there are no listeners attached, then none of those events will get processed and none of the logic after mousemove does anything, but those events will still be dutifully created and then silently forgotten.
There are some exceptions where a hot Observable can be made to appear like a cold Observable say by using replay, but that is simply buffering events for future subscribers, not changing the nature of the source.
As to your question regarding merging hot and cold. Since one of the sources is hot then the combined source would also be considered hot, because again it would continue to produce events regardless of the presence of subscribers. i.e.
//Still hot because the mouse does not stop generating events in the
//absence of subscribers.
var source = Rx.Observable.fromEvent(window, 'mousemove')
.merge(Rx.Observable.from([1,2,3,4,5]));