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 f_ R(r) of R.
Hint:. What is the PDF of the length of the smaller piece? For 0<r<1,
f_ R(r)=\quad
Find \mathbf{E}[R].
\mathbf{E}[R]=\quad
To find the PDF of R, we first need to determine the PDF of the length of the smaller piece, Y.
Since Y is uniformly distributed on [0,1], the PDF of Y can be written as:
f_Y(y) = 1 for 0 ≤ y ≤ 1
0 otherwise
Next, we need to find the cumulative distribution function (CDF) of Y, denoted as F_Y(y). The CDF of Y is given by:
F_Y(y) = ∫[0,y] f_Y(t) dt
Since the PDF of Y is constant within the range [0,1], the CDF of Y is simply:
F_Y(y) = y for 0 ≤ y ≤ 1
Now, we can determine the PDF of the length of the shorter piece, which is given by Y. The PDF of Y, denoted as f_Y(y), is simply the derivative of the CDF of Y:
f_Y(y) = d/dy F_Y(y)
Since F_Y(y) = y for 0 ≤ y ≤ 1, the derivative of F_Y(y) is simply:
f_Y(y) = 1 for 0 ≤ y ≤ 1
0 otherwise
Now, let's consider the ratio R = (length of shorter piece)/(length of longer piece).
We have two cases to consider:
1. If Y ≤ 0.5:
In this case, the length of the shorter piece is Y and the length of the longer piece is 1-Y. Therefore, the ratio R is:
R = (length of shorter piece)/(length of longer piece) = Y/(1-Y)
2. If Y > 0.5:
In this case, the length of the shorter piece is 1-Y and the length of the longer piece is Y. Therefore, the ratio R is:
R = (length of shorter piece)/(length of longer piece) = (1-Y)/Y
We can see that for 0 < r < 1, the ratio R can only take one of two possible values: r or 1/r.
Now, let's find the probability distribution function of R, f_R(r).
For 0 < r < 1:
f_R(r) = P(R = r) = P(Y/(1-Y) = r, where Y ≤ 0.5) + P((1-Y)/Y = r, where Y > 0.5)
Since Y is uniformly distributed on [0,1], the probability P(Y ≤ 0.5) = 0.5 and P(Y > 0.5) = 0.5.
Therefore, we can write the PDF of R as:
f_R(r) = P(R = r) = (0.5)(P(Y/(1-Y) = r) + P((1-Y)/Y = r))
Substituting the values for each case, we have:
f_R(r) = (0.5)(P(Y/(1-Y) = r) + P((1-Y)/Y = r))
= (0.5)(P(Y = r(1-Y)) + P(Y = (1-r)Y))
= (0.5)(P(Y = r - rY^2) + P(Y = Y - rY))
= (0.5)(f_Y(r - rY^2) + f_Y(Y - rY))
= (0.5)(1 + 1) = 1 for 0 < r < 1
Therefore, the PDF of R, f_R(r), is 1 for 0 < r < 1.
To find E[R], we need to calculate the expected value of R using the PDF of R. Since f_R(r) is equal to 1 for 0 < r < 1, the expected value of R can be calculated as:
E[R] = ∫[0,1] r * f_R(r) dr
= ∫[0,1] r * 1 dr
= [r^2/2] from 0 to 1
= 1/2 - 0
= 1/2
Therefore, the expected value of R, E[R], is 1/2.
To find the PDF of R, we need to find the cumulative distribution function (CDF) F_R(r) first.
Let's consider the case when Y ≤ r. In this case, the length of the shorter piece is Y, and the longer piece is 1 - Y. Since Y is uniformly distributed on [0,1], the probability of Y ≤ r is r. Therefore, the CDF can be written as:
F_R(r) = P(R ≤ r) = P(Y ≤ r) = r
Next, we can differentiate the CDF to obtain the PDF:
f_R(r) = d/d_r F_R(r) = d/d_r r = 1, for 0 < r < 1.
So, the PDF of R, denoted by f_R(r), is 1 for 0 < r < 1.
To find E[R], we need to calculate the expected value of R using the PDF:
E[R] = ∫ r * f_R(r) dr
Since f_R(r) is a constant of 1 for 0 < r < 1, the expected value can be written as:
E[R] = ∫ r * 1 dr from 0 to 1
E[R] = ∫ r dr from 0 to 1
E[R] = [r^2/2] from 0 to 1
E[R] = (1^2/2) - (0^2/2) = 1/2
Therefore, the expected value of R, denoted by E[R], is 1/2.