A stick of length 1 is broken into two pieces of length Y and 1−Y respectively, where Y is uniformly distributed on [0,1] . Let R be the ratio of the length of the shorter to the length of the longer piece.
Find the PDF fR(r) of R .
Hint:. What is the PDF of the length of the smaller piece? For 0<r<1,
fR(r)= ???
Find E[R] .
E[R]= ????
fR(r)= 2 / ((r + 1)^2)
1-2*(1-ln(2))
E(R)=ln(4)-1
1. fR(r)= 2/((r+1)^2)
2. E[R]= 0.38629436112
why is fR(r) = 2/(r+1)^2 and not 1/(r+1)^2 ?? where does the 2 com from ??
ah ok, i thing i got it, 1-Y is once the shorter end and once the longer
To find the PDF of R, we need to determine the probability density function (PDF) of the length of the smaller piece, Y.
Since Y is uniformly distributed on [0,1], the PDF of Y, denoted as fY(y), is constant over the interval [0,1].
To find the PDF of R, we can use the fact that the length of the shorter piece is Y, and the length of the longer piece is 1-Y.
Since R represents the ratio of the length of the shorter to the length of the longer piece, we have R = Y / (1-Y). Solving this equation for Y, we get Y = R / (1+R).
To find the PDF of R, fR(r), we can use the probability density transformation formula. The formula states that if Z = g(Y) is a function of Y, then the PDF of Z, denoted as fZ(z), is given by fZ(z) = |dfY(y)/dz|, where dfY(y)/dz is the derivative of the CDF of Y evaluated at y, with respect to z.
Using this formula, we can find the PDF of R as follows:
fR(r) = |dfY(y)/dr|, where y = r / (1+r)
First, let's find the derivative of the CDF of Y with respect to Y:
dFY(y)/dy = 1, since the CDF of a uniform distribution is a straight line with a unit slope.
Next, let's find the derivative of Y with respect to R:
dy/dr = 1 / (1+r)^2
Now, we can substitute these values into the formula:
fR(r) = |1 / (1+r)^2|
Simplifying this expression, we get:
fR(r) = 1 / (1+r)^2
This is the PDF of R for 0 < r < 1.
To find E[R], we can use the definition of expected value:
E[R] = ∫[0,1] r * fR(r) dr
Using the values of fR(r) from above, we can compute E[R] as follows:
E[R] = ∫[0,1] r * (1 / (1+r)^2) dr
This integral can be evaluated using integration techniques to find the expected value of R.
fR(r)=
2/(r+1)^2
submitted
2(r+1)2
Find E[R] .
E[R]=
0.23
submitted