Suppose that the random variable X takes on the values 0,1,. with the respective probabilities P[X = x]=. x= 0,1. Then: 1. Determine the constant c. 2. Compute the following probabilities: P[X23] P[X=2k+1 k-0,1.] P[X=3k +1, k =0,1.]
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- if x1, x2, ..., Xn be a random sample from Bin(4,0). is T = x UMVUE ? Justify your answer.An ordinary (fair) coin is tossed 3 times. Outcomes are thus triples of "heads" (h) and "tails" (t) which we write hth, ttt, etc. For each outcome, let N be the random variable counting the number of heads in each outcome. For example, if the outcome is hhh, then N (hhh) = = 3. Suppose that the random variable X is defined in terms of N as follows: X=6N-2N²-3. The values of X are given in the table below. Outcome hhh hth hht thh htt tth ttt tht Value of X-3 1 1 1 1 1 -3 1 Calculate the probabilities P (X=x) of the probability distribution of X. First, fill in the first row with the values of X. Then fill in the appropriate probabilities in the second row. Value X of X P(X=x) 0 0 0 00 XB5. Let X₁, X₂, ..., Xn be IID random variable with common expectation µ and common variance o², and let X = (X₁ + + X₂)/n be the mean of these random variables. We will be considering the random variable S² given by (a) By writing or otherwise, show that S² (b) Hence or otherwise, show that n S² = (x₁ - x)². = Ĺ(X₂ i=1 X₁ X = (X₁-μ) - (x-μ) = Σ(X; -μ)² - n(X - μ)². i=1 ES² = (n-1)0². You may use facts about X from the notes provided you state them clearly. (You may find it helpful to recognise some expectations as definitional formulas for variances, where appropriate.) (c) At the beginning of this module, we defined the sample variance of the values x₁, x2,...,xn to be S = 1 n-1 n i=1 ((x₁ - x)². Explain one reason why we might consider it appropriate to use 1/(n-1) as the factor at the beginning of this expression, rather than simply 1/n. B6. (New) Roughly how many times should I toss a coin for there to be a 95% chance that between 49% and 510/ of my nain toon land Honda?
- 3. Let X be the random variable that takes on the integers {0, 1, 2, ..., 15} with equal probabilities. Define a new random variable Y = X + A, where A is a random variable that takes on the values {-1, 0, 1} with equal probabilities. If the RVs X and A are independent, find the mutual information between X and Y.A deck has only 51 cards left, because a spade has been removed. From this deck, cards will be drawn at random, in succession, without replacement. Let Xi be a random variable representing the number of spades in the future i th draw. Calculate the following, and present your answer with 5 digits after the decimal point: Prob(X1 + ... + X7 = 3) = [1]You are allowed to take a certain test three times, and your final score will be the maximum of the test scores. Your score in test k, where k = 1, 2, 3, takes one of the values from k to 10 with equal probability 1/(11 − k), independently of the scores in other tests. What is the PMF of the final score?
- 2. Let the independent random variables X1 and X2 have Bin(0.1,2) and Bin(0.5, 3), respectively. (a) Find P(X1 = 2 and X2 = 2). (b) Find P(X1 + X2 = 1). (c) Find E(X1 + X2). (d) Find Var(X1 + X2).Suppose a and b be two possible values of a random variable X with a > b. The probability that X lies between a and b is P(a > X > b) = F (a) - F (b) Select one: O True O FalseSuppose X1,X2.X10 are independent N(H,02) random variables. Z1,Z2.Z5 are independent N(0,1) random variables. The distribution of X + Z is a. N(u, O b. None of these С. N(u,- 10 d. N(0,
- Prob ... solve Q 1 & Q 2 please ?3.Suppose that X and Y are independent random variables for which Var(X)=Var(Y)=3. Find the values of (a) Var(X-Y); (b) Var(2X-3Y+1).Each of the random variables X and Y takes only 3 values {1,2,3} with the following probabilities: 1 1 0 1/6 1/6 y 2 1/6 0 1/6 3 1/6 1/6 0 2.