A fair coin is flipped independently until the first Heads is observed. Let K be the number of Tails observed before the first Heads (note that K is a random variable). For k=0,1,2,…,K, let Xk be a continuous random variable that is uniform over the interval [0,3]. The Xk's are independent of one another and of the coin flips. Let the random variable X be defined as the sum of all the Xk's generated before the first Heads. That is, X=∑Kk=0Xk. Find the mean and variance of X. You may use the fact that the mean and variance of a geometric random variable with parameter p are 1/p and (1−p)/p2, respectively.
E[X]= - unanswered
var(X)=
I ma stuck with the same problem, somebody please give us a hint.
To find the mean and variance of X, we can use the properties of expected value and variance.
Let's start with the expected value, E[X]. Since X is defined as the sum of the Xk's, we can use the linearity of expected value to find E[X].
E[X] = E[∑Kk=0Xk]
Since the Xk's are independent of each other, we can write this as:
E[X] = ∑Kk=0E[Xk]
Now, each Xk is a continuous random variable that is uniform over the interval [0,3]. The mean of a uniform distribution is the average of the lower and upper bounds, which in this case is (0+3)/2 = 1.5.
Therefore, E[Xk] = 1.5 for all k.
Now, we need to find the expected value of K, which represents the number of Tails observed before the first Heads. K follows a geometric distribution with parameter p = 1/2, since the coin is fair and the probability of getting Heads on each flip is 1/2.
The mean of a geometric distribution with parameter p is 1/p. So, E[K] = 1/(1/2) = 2.
Since K represents the number of Tails observed before the first Heads, the total number of Xk's generated before the first Heads is K + 1.
Therefore, the expected value of X is:
E[X] = ∑Kk=0E[Xk] = (K + 1) * E[Xk] = (2 + 1) * 1.5 = 4.5.
So, the mean of X is 4.5.
Now, let's move on to finding the variance of X, var(X).
Since the Xk's are independent of each other, we can use the property that the variance of the sum of independent random variables is equal to the sum of their variances.
var(X) = var(∑Kk=0Xk)
Again, using the linearity of variance, we can write this as:
var(X) = ∑Kk=0var(Xk)
Each Xk is a continuous random variable that is uniform over the interval [0,3]. The variance of a continuous uniform distribution over [a,b] is (b - a)^2/12.
In this case, a = 0 and b = 3, so the variance of each Xk is (3 - 0)^2/12 = 3/4.
Therefore, var(Xk) = 3/4 for all k.
Since K follows a geometric distribution, the variance of a geometric distribution with parameter p is (1−p)/p^2.
Therefore, the variance of K is var(K) = (1−1/2)/(1/2)^2 = 1.
Now, the total number of Xk's generated before the first Heads is K + 1, so the variance of X is:
var(X) = ∑Kk=0var(Xk) = (K + 1) * var(Xk) = (2 + 1) * 3/4 = 9/4.
So, the variance of X is 9/4.
In summary:
- The mean of X is 4.5.
- The variance of X is 9/4.
The E[Y] is equal to E[N]*E[X], being Y=X1+...+Xn. Now, we know that N is the number of coin flips, and Xk is independent and uniformly distributed between 0 and 3. This means that E[X]=3*1/2=3/2, with a variance of 3^2/12=3/4.
Now, for N we are given 1/p and the variance. Since it's a fair coin, p=1/2, and so E[n]=2 and var[N]=2.
With this, E[Y]=2*3/2=3
var(Y)=E[N]*var(X)+(E[X])^2*var(N)=2*3/2+2^2*3/4=3+3=6
Hope it's understandable.
E[X]= 3
var(X)= 6
if you are reading this answer, please help us with the whole problem set....