# pdf question

Discussion in 'Probability' started by B, Jul 2, 2010.

1. ### BGuest

Have:
X = requests per second with E[X]
Y = resources used per request with E[Y]
They are independent.
f(x) is probability density function of E[X]
g(y) is probability density function of E[Y]

Question:
How can h(f(x),g(y)) pdf of E[X*Y] be derived from f(x) and g(y)? For
example, for E[X+Y] h(x,y) is a convolution of f(x) and g(y).

In the end what I am really looking for is: given the pdf for arrivals
and pdf of resources per arrival, how can I easily/quickly get the
additional combined pdf/cdf s.t. I can say with 95% confidence that
the total resources consumed by all arrivals is less than x. FWIW, I
am working in the discrete world.

B, Jul 2, 2010

2. ### Bastian ErdnuessGuest

f is pdf of X, not of E[X]
here the same
again, h is the pdf of X*Y, not E[X*Y]. And h depends on one parameter,
not on two.
Here it is similar. For the sum you have

h(t) = pdf[X+Y](t) = int[0..t] f(t-s) g(s) ds

and for the product

h(t) = pdf[X*Y](t) = int[0..oo] f(t/s) g(s) ds

For the expectation you have E[X*Y] = E[X]*E[Y].
For the cdf you have

cdf[X*Y](t) = int[0..oo] cdf[X](t/s) pdf[Y](s) ds

Of course by the symmetry, you can also exchange the roles of X and Y if
this makes it easyer to calculate.

In the discrete case you have

P(X*Y <= n) = sum[k=1..oo] P(X <= n/k) P(Y = k)

Here, you get a finite sum, if P(X = 0) = 0 holds. If not, but
P(Y=0) = 0 holds, exchange the roles of X and Y. And

P(X*Y = n) = sum[k|n] P(X = n/k) P(Y = k) .

Cheers
Bastian

Bastian Erdnuess, Jul 2, 2010

3. ### HenryGuest

You can actually deal with the zero case as

P(X*Y=0) = P(X=0) + P(Y=0) - P(X=0,Y=0)
= P(X=0) + P(Y=0) - P(X=0) P(Y=0)

and

P(X*Y <= n) = P(Y=0) + sum[k=1..oo] P(X <= n/k) P(Y = k)

Henry, Jul 2, 2010
4. ### danheymanGuest

What you DON'T want is the distribution of X*Y. Let Y_i be the
resources needed by the ith arrival. The total number of resources
needed is S=Y_1 +Y_2 +...+Y_X, where the Y's have density g. Assuming
the Y's are iid and independent of X, the probability generating
function (since you say the RVs are discrete) of the distribution of S
is easily obtained in terms of PGFs of f and g (look up compound
distribution). Numerical inversion may be needed, but in some special
cases, the PGF can be inverted analytically (X is Poisson and Ys are
Poisson or geometric if memory serves).

danheyman, Jul 7, 2010