Probability Density Function of two Resistors in Parallel

[darthhath]

I have a problem where there are two resistors in parallel and I need to find the equivalent resistance. R1 = X and R2 = Y, and X and Y are independent random variables, uniform over the range of 100-120.

If R equivalent = Z = XY/X+Y, what is probability density function of Z?

 

[juvenal]

I played around with this in Mathematica, but didn't get it to work. But I don't have time to check my work or play around with this much more.

What I did was a change of variables:

u= [(1/x) + (1/y)]^-1 = What you call z
v = x

Therefore:

x = v
y = uv/(v-u)

The limits of integration are now:

u: 50 to 60
v: 100u/(100-u) to 120u/(120-u)
(This may be where I messed up)

Now, f(u,v) = f(x,y)*|J| where J is the Jacobian of the transformation and
f(x,y) = 1/400 inside the ranges of x and y, 0 outside.

Then f(u) = Integral of f(u,v) with respect to v.

 

[darthhath]

I am unclear on everything except for the limits of integration. I know that you may have messed up, but if you could let me know what you are shooting for, maybe I can figure them out.

Thanks

 

[juvenal]

Well, I did the integration of my joint pdf over all of u and v, and did not get 1. So my answer is wrong.

The whole point is that you want to transform from x,y to u(=z). However, I don't know how to do that - transform two variables to one. However, you can transform two variables to two variables, and integrate out one of the variables, which in this case is v. Once you integrate over v, then you will be left with the pdf of u, which is what you want.

The transformation itself requires calculating the absolute value of the Jacobian, as well as evaluating the new limits.

Unfortunately, this is as much detail as I am willing to provide, so I would recommend looking at a decent statistics textbook like Casella and Berger.

 

[OlderDan]

I played around with this in Mathematica, but didn't get it to work. But I don't have time to check my work or play around with this much more.

What I did was a change of variables:

u= [(1/x) + (1/y)]^-1 = What you call z
v = x

Therefore:

x = v
y = uv/(v-u)

The limits of integration are now:

u: 50 to 60
v: 100u/(100-u) to 120u/(120-u)
(This may be where I messed up)

Now, f(u,v) = f(x,y)*|J| where J is the Jacobian of the transformation and
f(x,y) = 1/400 inside the ranges of x and y, 0 outside.

Then f(u) = Integral of f(u,v) with respect to v.

v: 100u/(100-u) to 120u/(120-u)
(This may be where I messed up)

I think you are right. If you think in terms of the reciprocals of the resistance, for every Z there is a range of possible X values, with a unique Y corresponding to each X. At the limiting values of 1/Z (1/60 and 1/50) there is only 1 value of 1/X (1/120 and 1/100). Starting at 1/Z = 1/60 and increasing (decreasing Z), all 1/X values from 1/120 to a maximum are permitted, with the maximum being 1/Z - 1/120. So the range of possible values of 1/X increases linearly with 1/Z until the maximum 1/X reaches 1/100. Any additional increase in 1/Z places a lower limit on the value of 1/X, with a maxiumum 1/X at 1/100. The lower limit is also linear in 1/Z, so the range of 1/X shrinks linearly toward zero.

The point where the X range is maximized is not in the middle of the Z range, but it is in the middle of the 1/Z range. The middle of the 1/Z range occurs at

Z = 600/11 = 54.545454

The 1/Z distribution is not quite triangular, because the 1/X distribution is not quite uniform, and I'm not sure if changing back to the Z distribution from 1/Z is going to "undue" the non-linearity or compouund it, probably the latter. Roughly speaking the Z distributions should be approximately triangular from zero at 50 and 60, peaking at 54.5454. Maybe you want to revise your approach and get the final answer.