Probability

P={ z + x*y/(y+a) > x/(x*k+1)}
where x,y,z are exponentially distributed with mean xbar,ybar,zbar.
a,k are constants.
please suggest me how to evaluate this probability.....
for k=0, P lies between 0 and 1 but for any positive value of k, P become higher than one which is unacceptable....
take xbar=1 ybar=3.14 zbar=1

GW008 said:

P={ z + x*y/(y+a) > x/(x*k+1)}
where x,y,z are exponentially distributed with mean xbar,ybar,zbar.
a,k are constants.
please suggest me how to evaluate this probability.....
for k=0, P lies between 0 and 1 but for any positive value of k, P become higher than one which is unacceptable....
take xbar=1 ybar=3.14 zbar=1

Click to expand...

I can't imagine finding a closed form for this probability! And, with two parameters (a, k), I wouldn't want to do the calculations to find it by Monte Carlo.

You should notice that the given expression is NOT the probability - rather, the probability is the likilhood that the expression would be true. I would write it as

\(\displaystyle F(x,y,z; a,k) = z + \dfrac{x\ y}{y+a} - \dfrac{x}{k\ x +1} > 0\)

This expression may well exceed 1, but the probability will ALWAYS be < 1.

You probably know (or it is your notes) that for an exponential distribution with mean \(\displaystyle \bar x\), the Variance is \(\displaystyle \bar x^2\). Can you find the expectation value, \(\displaystyle \mathrm E[F]\)? How about the Variance, \(\displaystyle \mathrm V[F]\)? That is much harder because you have to use propagation of errors to combine properly the Variances of (x, y, z).

What did the question really ask for?