A fair coin is flipped independently until the first Heads is observed. Let the random variable K be the number of tosses until the first Heads is observed plus 1. For example, if we see TTTHTH, then K= 5. For k = 1,2,...,K, let Xk be a continuous random variable that is uniform over the interval [0, 5]. The Xkare independent of one another and of the coin flips. Let X=sum(Xk). Find the mean and variance of X.
Firstly, I assume N=K in your solutions. The expected value and variance of X can be found via Law of Iterated Expectation (LIE) and Law of Total Variance (LTV):
E[X]=E[E[X|K]], var(X)=E[var(X|K)]+var(E[X|K])
For the expectation, your approach is correct, but it can be found via LIE:
E[X|K]=KE[Xk]→E[KE[Xk]]=E[K]E[Xk]
You just need to correct your expectation for K: E[K]=1/p+1, since it is of the form 1+Y, where Y is a geometric RV with parameter p. Also, note that var(K)=var(1+Y)=var(Y)=(1−p)/p2 as yours.
For the variance, we need var(X|K)=var(∑Xk|K)=Kvar(Xk), and by LTV:
var(X)=E[Kvar(Xk)]+var(KE[Xk])=var(Xk)E[K]+E[Xk]2var(K)
Substituting:
E[X] = 15/2
Var[X] = 18.75
I have already posted this problem before 6 days, and never got any response. Please Help.
To find the mean and variance of X, we need to calculate the mean and variance of each Xk, and then sum them up.
1. Mean of Xk:
Since Xk is a continuous random variable that is uniformly distributed over the interval [0, 5], the mean can be calculated using the formula for the mean of a uniform distribution:
Mean(Xk) = (a + b) / 2 = (0 + 5) / 2 = 2.5
2. Variance of Xk:
Similarly, the variance of a continuous uniform distribution can be calculated using the formula:
Variance(Xk) = (b - a)^2 / 12 = (5 - 0)^2 / 12 = 25 / 12 ≈ 2.0833
3. Sum of Xk:
Since X is defined as the sum of Xk for k = 1, 2, ..., K, where K is the number of tosses until the first Heads is observed plus 1, we can calculate the mean of X by summing up the means of all Xk. Since Xk is independent of the coin flips, its mean is constant regardless of the number of tosses.
Mean(X) = K * Mean(Xk) = K * 2.5
Similarly, the variance of X can be calculated by summing up the variances of all Xk.
Variance(X) = K * Variance(Xk) = K * 2.0833
However, K is a random variable representing the number of tosses until the first Heads is observed plus 1, so we need to find the mean and variance of K first.
4. Mean of K:
The mean of K can be calculated using the geometric distribution formula:
Mean(K) = 1 / p = 1 / 0.5 = 2
5. Variance of K:
The variance of K for a geometric distribution is given by:
Variance(K) = (1 - p) / (p^2) = (1 - 0.5) / (0.5^2) = 1 / 0.25 = 4
Now, substituting the mean and variance of K back into the mean and variance formulas for X, we get:
Mean(X) = 2 * 2.5 = 5
Variance(X) = 2 * 2.0833 = 4.1666
Therefore, the mean of X is 5 and the variance of X is approximately 4.1666.
To find the mean and variance of X, we first need to find the probability distribution of K, the number of coin tosses until the first heads.
The probability distribution of K can be modeled as a geometric distribution, where each toss of the fair coin has a probability of success (getting a heads) of 1/2.
The probability of getting the first heads on the k-th toss is given by: P(K = k) = (1/2)^(k-1) * (1/2) = (1/2)^k.
Now, let's find the mean of X, denoted as E(X):
E(X) = E(X1 + X2 + ... + XK)
Since we know that X1, X2, ..., XK are all independent and uniformly distributed between 0 and 5, we can use the linearity of expectation to simplify the calculation.
E(X) = E(X1) + E(X2) + ... + E(XK)
Since each Xi is uniformly distributed between 0 and 5, the expected value of each Xi is the midpoint of this interval, which is (0 + 5) / 2 = 2.5.
E(X) = 2.5 + 2.5 + ... + 2.5 (K times)
E(X) = 2.5 * K
Now, let's find the variance of X, denoted as Var(X):
Since X1, X2, ..., XK are all independent, we can use the property that Var(aX + b) = a^2 * Var(X), where a and b are constants.
Var(X) = Var(X1 + X2 + ... + XK)
Var(X) = Var(X1) + Var(X2) + ... + Var(XK)
Since each Xi follows a uniform distribution between 0 and 5, the variance of each Xi is given by (5 - 0)^2 / 12 = 25 / 12.
Var(X) = 25 / 12 + 25 / 12 + ... + 25 / 12 (K times)
Var(X) = K * (25 / 12)
Therefore, the mean of X is 2.5K and the variance of X is 25K/12.