Efficient way to calculate the "product" of a discrete convolution

Question

I'm looking for an elegant way to compute the "product" of a discrete convolution, instead of the sum.

Here is the formula of a discrete convolution:

In this case we can use: conv(x,y)

Now I would like to implement those operations

Of course I can use a loop, but I'm looking for a trick in order to linearize this operation.

EXAMPLE:

f = [2 4 3 9 7 1]
g = [3 2 1]

dist = length(g)-1;

for ii = 1:length(f)-dist
    x(ii) = prod(f(ii:ii+dist).*g)
end

x =

144    648   1134    378

Answer 1

Another solution partly inspired by Dev-iL answer to relatively the same question

exp(sum(log(g))+conv(log(f),[1 1 1],'valid'))

or

exp(sum(log(g))+movsum(log(f),numel(g),'Endpoints', 'discard'))

since exp(sum(log(x))) = prod(x)

But here instead of one vector we have two vectors f and g.

The desired formula can be reformulated as:

Timing in octave:

f= rand(1,1000000);
g= rand(1,100);

disp('----------EXP(LOG)------')
tic
    exp(sum(log(g))+conv(log(f),ones(1,numel(g))));
toc

disp('----------BSXFUN------')
tic
    ind = bsxfun(@plus, 0:numel(f)-numel(g), (1:numel(g)).');
    x = prod(bsxfun(@times, f(ind), g(:)),1);
toc
disp('----------HANKEL------')
tic
    prod(g)*prod(hankel(f(1:numel(g)), f(numel(g):end)));
toc

disp('----------CUMPROD-----')
tic
    pf = cumprod(f);
    x = prod(g)*pf(numel(g):end)./[1 pf(1:(end-numel(g)))];
toc

Result:

----------EXP(LOG)------%rahnema1
Elapsed time is 0.211445 seconds.
----------BSXFUN--------%Luis Mendo
Elapsed time is 1.94182 seconds.
----------HANKEL--------%gnovice
Elapsed time is 1.46593 seconds.
----------CUMPROD-------%gnovice
Elapsed time is 0.00748992 seconds.