In the Bullish window put...

Sum(C>Mov(C,50,E),3)=3;

and in the Bearish window put...

Sum(C<Mov(C,50,E),3)=3;

Roy

Hi J

Perhaps if you'd tried my sugestion you would have found that it does work. How do I know this? Two reasons. Not only did I try it to see that it was correct, but I also have the advantage of knowing that an EMA ALWAYS moves toward the price. Therefore, if the price is above the EMA for three consecutive bars it follows that the EMA is moving up for three consecutive bars. Which is what you asked for isn't it?

Your suspicions about my solution would have been well founded if you'd asked about any other type of MA but you were specific about it being an exponential MA. Wilders (being exponential in nature) also follows this rule (always moving toward the price), and I see no reason to think that DEMA, TEMA or any other exponential MA variation would behave any differently.

Here's how you can simulate a standard exponential moving average function with MFL code. Essentially the value of "Rate" sets the proportion of old data (as calculated on the previous bar) to new data when calculating the new EMA value. If you think about it you'll realize that in this example there's an automatic bias of any new EMA value towards the data array being smoothed - and this is irrespective of the number of periods used. I hope this helps but don't just take my word for it - check it out for yourself.

{Exponential Moving Average}
Pds:=Input("Periods",1,1000,10);
Rate:=2 / (Pds+1);
If(Cum(1)=1, C, PREV*(1-Rate) + C*Rate );

Roy

Hi J

If you still need convincing that my suggested method offers a genuine solution to your problem then create an indicator with the following code and drop it onto any chart. Look for any differences in the results - there won't be any unless the MA method is changed.

{Compare 3 methods}
M:=Mov(C,50,E);
{1} Sum(C>M,3)=3;
{2} Sum(M>Ref(M,-1),3)=3;
{3} M>Ref(M,-1) AND Ref(M,-1)>Ref(M,-2) AND Ref(M,-2)>Ref(M,-3);

Roy