Prevent onmouseout when hovering child element of the parent absolute div WITHOUT jQuery

Use onmouseleave.

Or, in jQuery, use mouseleave()

It is the exact thing you are looking for. Example:

<div class="outer" onmouseleave="yourFunction()">
    <div class="inner">
    </div>
</div>

or, in jQuery:

$(".outer").mouseleave(function(){
    //your code here
});

an example is here.

For a simpler pure CSS solution that works in most cases, one could remove children's pointer-events by setting them to none

.parent * {
     pointer-events: none;
}

Browser support: IE11+

function onMouseOut(event) {
        //this is the original element the event handler was assigned to
        var e = event.toElement || event.relatedTarget;
        if (e.parentNode == this || e == this) {
           return;
        }
    alert('MouseOut');
    // handle mouse event here!
}



document.getElementById('parent').addEventListener('mouseout',onMouseOut,true);

I made a quick JsFiddle demo, with all the CSS and HTML needed, check it out...

EDIT FIXED link for cross-browser support http://jsfiddle.net/RH3tA/9/

NOTE that this only checks the immediate parent, if the parent div had nested children then you have to somehow traverse through the elements parents looking for the "Orginal element"

EDIT example for nested children

EDIT Fixed for hopefully cross-browser

function makeMouseOutFn(elem){
    var list = traverseChildren(elem);
    return function onMouseOut(event) {
        var e = event.toElement || event.relatedTarget;
        if (!!~list.indexOf(e)) {
            return;
        }
        alert('MouseOut');
        // handle mouse event here!
    };
}

//using closure to cache all child elements
var parent = document.getElementById("parent");
parent.addEventListener('mouseout',makeMouseOutFn(parent),true);

//quick and dirty DFS children traversal, 
function traverseChildren(elem){
    var children = [];
    var q = [];
    q.push(elem);
    while (q.length > 0) {
      var elem = q.pop();
      children.push(elem);
      pushAll(elem.children);
    }
    function pushAll(elemArray){
      for(var i=0; i < elemArray.length; i++) {
        q.push(elemArray[i]);
      }
    }
    return children;
}

And a new JSFiddle, EDIT updated link