1.Let 𝑋 and 𝑌 be two binomial random variables:
a.If 𝑋 and 𝑌 are independent, then 𝑋+𝑌 is also a binomial random variable
b.If 𝑋 and 𝑌 have the same parameters, 𝑛 and 𝑝 , then 𝑋+𝑌 is a binomial random variable
c.If 𝑋 and 𝑌 have the same parameter 𝑝 , and are independent, then 𝑋+𝑌 is a binomial random variable.
2.Suppose that, 𝐄[𝑋]=0 . Then, 𝑋=0 .
3.Suppose that, 𝐄[𝑋^2]=0 . Then, 𝐏(𝑋=0)=1 .
4.Let 𝑋 be a random variable. Then, 𝐄[𝑋^2]≥𝐄[𝑋]. True or false?
5.Suppose that, 𝑋 is a random variable, taking positive integer values, which satisfies 𝐄[(𝑋−6)^2]=0 . Then, 𝑝𝑋(4)=𝑝𝑋(5) .
6.Suppose that 𝐄[𝑋]≥0 . Then, 𝑋≥0 with probability 1, i.e., 𝐏(𝑋≥0)=1.
True or false?
1.a False
1.b False
1.c True
2. False
3. True
4. False
5. True
6. False
1.
a) False
b) False
c) True
2. False
3. True
1.a. Yes, when X and Y are independent, X+Y is also a binomial random variable. But be careful, the jokes in binomial distribution are "binomial" hilarious!
b. Yes, when X and Y have the same parameters, X+Y is indeed a binomial random variable. They're like binomial twins!
c. Absolutely! When X and Y are independent and have the same parameter, X+Y becomes a binomial random variable. It's binomially fantastic!
2. Ah, if E[X]=0, it doesn't necessarily mean X=0. X could be hiding behind the mathematical curtains!
3. Hold on there! Just because E[X^2]=0 doesn't automatically mean P(X=0)=1. X can be a mysterious little rascal!
4. False! E[X^2] can be greater or lesser than E[X]. It's like comparing apples and oranges, or in this case, expectations.
5. Hmm, with E[(X-6)^2]=0, it might not lead to P(X=4)=P(X=5). X seems to have a mind of its own there!
6. Oh, no no! Just because E[X] is greater than or equal to 0, doesn't guarantee that X is always greater than or equal to 0. Expectations can be tricky like that!
1. To determine whether the statements are true or false, we need to understand the properties of binomial random variables. A binomial random variable is defined by two parameters, n (number of trials) and p (probability of success in each trial). The random variable X represents the number of successes in n independent Bernoulli trials.
a. If X and Y are independent binomial random variables, then X+Y represents the total number of successes in the combined trials. Since X and Y are independent, the trials do not affect each other, and the probability of success in each trial remains the same. Therefore, X+Y is still a binomial random variable.
b. If X and Y have the same parameters n and p, it means they are defined by the same number of trials and the same probability of success in each trial. In this case, X+Y represents the total number of successes in the combined trials, which still follows the same parameters n and p. Hence, X+Y is a binomial random variable.
c. If X and Y have the same parameter p and are independent, it means they have the same probability of success in each trial, but the number of trials can be different. In this case, X+Y represents the total number of successes in the combined trials, and the probability of success remains the same p. Therefore, X+Y is a binomial random variable.
Therefore, all three statements are true.
2. The statement "E[X] = 0 implies X = 0" is false. The expected value (mean) of a random variable measures the average value it takes, but it does not necessarily mean that the random variable itself is always equal to that mean value. In this case, E[X] = 0 means that the expected value of X is zero, but it does not imply that X is always equal to zero. X can still take other values, including negative ones.
3. The statement "E[X^2] = 0 implies P(X = 0) = 1" is true. The expected value of X^2 being zero means that the square of X has an average value of zero. Since the square of a variable can never be negative, it implies that X must always be zero. Therefore, P(X = 0) = 1, meaning that the probability of X being zero is equal to 1.
4. The statement "E[X^2] ≥ E[X]" is true. The expected value of X^2 is generally greater than or equal to the expected value of X. This inequality holds because squaring a number usually results in a larger value, and when averaging those squared values, the result tends to be greater than or equal to the average of the original values.
5. The statement "E[(X-6)^2] = 0 implies P(X = 4) = P(X = 5)" is true. If the expected value of (X-6)^2 is zero, it implies that (X-6) must always be equal to zero. This means X must always be equal to 6. Since X is a positive integer and can only take integer values, the probability of X being equal to 4 or 5 is 0, and therefore P(X = 4) = P(X = 5) = 0.
6. The statement "E[X] ≥ 0 implies X ≥ 0 with probability 1" is true. If the expected value of X is greater than or equal to zero, it means that, on average, X is a non-negative value. This implies that X itself is non-negative with probability 1, meaning that the probability of X being greater than or equal to zero is equal to 1.