Let k be a positive integer and let X be a continuous random variable that is uniformly distributed on [0,k]. For any number x, denote by [x] the largest integer not exceeding x. Similarly, denote fraction(x)=x-[x] to be the fractional part of x. The following are two properties of [x] and fractions(x):
x=[x]+frac(x)
[x]<=x<[x]+1
frac(x) element of (0,1).
For example, if x=2.91, then [x]=2 and frac(x)=0.91.
1. Let Y=[x] and let p_Y (y) be its PMF. There exists some non negative integer l such that p_Y (y) >0 for every y element of {0,1,.......,l} and p_Y (y)=0 for y>=l+1. Find l and p_Y (y) for y element of {0,1,......,l}. Your answer should be a function of k.
2. Let Z =frac(x) and let f_Z (z) be its PDF. There exists a real number c such that f_Z (z)>0 for every z element of (0,c), and f_Z (z)=0 for every z>c. Find c, and f_Z (z) for z element of (0,c).
1. Since X is uniformly distributed on [0,k], the value of Y=[X] will be an integer between 0 and k. Let's consider the values of Y for each interval of [0,k].
If 0 ≤ X < 1, then Y = 0.
If 1 ≤ X < 2, then Y = 1.
If 2 ≤ X < 3, then Y = 2.
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.
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If k-1 ≤ X < k, then Y = k-1.
From these intervals, we can see that the non-negative integer l is equal to [k]. Therefore, l = [k].
Now, let's find p_Y(y) for y ∈ {0,1,...,l}.
For y = 0, the probability p_Y(0) can be calculated as the probability that X falls between 0 and 1:
p_Y(0) = P(0 ≤ X < 1) = 1/k.
For y = 1, the probability p_Y(1) can be calculated as the probability that X falls between 1 and 2:
p_Y(1) = P(1 ≤ X < 2) = 1/k.
Similarly, for y = 2, the probability p_Y(2) can be calculated as the probability that X falls between 2 and 3:
p_Y(2) = P(2 ≤ X < 3) = 1/k.
This pattern continues until y = l. Therefore, for y = 0, 1, 2, ..., l:
p_Y(y) = 1/k.
2. The fractional part of X, Z = frac(X), falls between 0 and 1. Using the fact that X is uniformly distributed on [0,k], we can find the probability density function (PDF), f_Z(z), for z ∈ (0,1).
Since Z = frac(X) = X - [X], we can rewrite the PDF as:
f_Z(z) = f_X(z + [X]).
For z ∈ (0,1), the minimum possible value of [X] is 0 and the maximum possible value of [X] is l, as found in question 1. Therefore, we can write the PDF as:
f_Z(z) = f_X(z) + f_X(z + 1) + f_X(z + 2) + ... + f_X(z + l),
where f_X(z) is the PDF of X.
Since X is uniformly distributed on [0,k], the PDF of X, f_X(z), is a constant 1/k for z ∈ [0,k] and 0 otherwise.
Therefore, for z ∈ (0,1):
f_Z(z) = 1/k + 1/k + ... + 1/k (l times) = l/k.
To find the value of c, we need to determine the upper limit of z for which f_Z(z) is positive. Since f_Z(z) = 0 for z > c, we can find c by setting f_Z(c) = 0:
l/k = 0.
This implies that l = 0.
Therefore, for z ∈ (0,1):
f_Z(z) = 0.
We can conclude that c = 0 and f_Z(z) = 0 for z ∈ (0,1).
1. To find the value of l, we note that Y takes on the values of the largest integer not exceeding x. Since x is uniformly distributed on the interval [0, k], the maximum value of Y would be [k]. However, since [k] is not an integer, we need to take the largest integer smaller than [k]. Hence, l = [k] - 1.
Now, let's find p_Y(y) for y ∈ {0, 1, ..., l}. Since Y is the largest integer not exceeding x, p_Y(y) represents the probability that Y takes on the value y.
For y = 0, we have p_Y(0) = P(Y = 0) = P(0 ≤ x < 1) = 1/k.
For y = 1, we have p_Y(1) = P(Y = 1) = P(1 ≤ x < 2) = 1/k.
For y = 2, we have p_Y(2) = P(Y = 2) = P(2 ≤ x < 3) = 1/k.
...
For y = l, we have p_Y(l) = P(Y = l) = P(l ≤ x < l+1) = 1/k.
So, for y ∈ {0, 1, ..., l}, p_Y(y) = 1/k.
2. To find the value of c, we note that Z is defined as the fractional part of x, which ranges from 0 to 1. Hence, c = 1.
Now, let's find f_Z(z) for z ∈ (0, c). Since Z represents the fractional part of x, f_Z(z) represents the probability density function of Z.
For z ∈ (0, 1), we have f_Z(z) = P(Z = z) = P(z ≤ frac(x) < z+dz), where dz is a small interval around z.
Since X is uniformly distributed on [0, k], the length of the interval [0, k] is k − 0 = k. So, the length of the interval [0, 1] is 1.
Therefore, f_Z(z) should be constant within the interval (0, 1) to maintain a uniform distribution.
Hence, for z ∈ (0, 1), f_Z(z) = 1.
Note: It is important to clarify whether the problem is asking for the probability mass function (PMF) or the probability density function (PDF) of Y and Z. The solution assumes that Y and Z are discrete and continuous random variables, respectively.