Let \theta \sim \pi (\theta ) be a parameter supported on \mathbb {Z}, the integers. Suppose that we observe random variables
Y_ i=\theta X_ i
for i = 1, \ldots , n. The outcomes of X_1, \ldots , X_ n are unknown to you, but you do know that they are i.i.d. and uniformly distributed on the set \{ -1,0,1\}. Assume that X_ i is independent of \theta for all i.
Compute each of the probabilities below:
\mathbb {P}(Y_1=6 \text { and } Y_2=0|\theta =3)=
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\mathbb {P}(Y_1=7 \text { and } Y_2=-7 \text { and } Y_3\in \{ 0,7\} |\theta = -7)=.
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\mathbb {P}(Y_1=Y_2+Y_3|\theta =5).
To compute each of the probabilities, we need to use the definition of the random variable Y_i and the properties of the given distributions.
1. We have Y_i = theta * X_i. If theta = 3, then we can substitute this value into Y_i = theta * X_i to get:
Y_i = 3 * X_i.
Now, we are given Y_1 = 6 and Y_2 = 0. We want to find the probability of this event given theta = 3:
P(Y_1 = 6 and Y_2 = 0 | theta = 3).
Using the definition of Y_i, we can rewrite this probability as:
P(3 * X_1 = 6 and 3 * X_2 = 0 | theta = 3).
Since X_1 and X_2 are independent and uniformly distributed on {-1, 0, 1}, we can compute the probabilities of each individual event:
P(X_1 = 2) = P(X_1 = -2) = 0
P(X_1 = 0) = 1/3
P(X_2 = 0) = 1/3
Now, since X_1 and X_2 are independent, we can multiply the probabilities:
P(X_1 = 0 and X_2 = 0) = P(X_1 = 0) * P(X_2 = 0) = (1/3) * (1/3) = 1/9
Therefore, the probability P(Y_1 = 6 and Y_2 = 0 | theta = 3) is equal to the probability of X_1 = 0 and X_2 = 0. So the answer is 1/9.
2. We have Y_i = theta * X_i. If theta = -7, then we can substitute this value into Y_i = theta * X_i to get:
Y_i = -7 * X_i.
Now, we are given Y_1 = 7, Y_2 = -7, and Y_3 ∈ {0, 7}. We want to find the probability of this event given theta = -7:
P(Y_1 = 7 and Y_2 = -7 and Y_3 ∈ {0, 7} | theta = -7).
Using the definition of Y_i, we can rewrite this probability as:
P(-7 * X_1 = 7 and -7 * X_2 = -7 and -7 * X_3 ∈ {0, 7} | theta = -7).
Since X_1, X_2, and X_3 are independent and uniformly distributed on {-1, 0, 1}, we can compute the probabilities of each individual event:
P(X_1 = -1) = P(X_1 = 1) = 1/3
P(X_1 = 0) = 1/3
Now, since X_1, X_2, and X_3 are independent, we can multiply the probabilities:
P(X_1 = -1 and X_2 = 1 and X_3 = 0) = P(X_1 = -1) * P(X_2 = 1) * P(X_3 = 0) = (1/3) * (1/3) * (1/3) = 1/27
Therefore, the probability P(Y_1 = 7 and Y_2 = -7 and Y_3 ∈ {0, 7} | theta = -7) is equal to the probability of X_1 = -1, X_2 = 1, and X_3 = 0. So the answer is 1/27.
3. We want to find the probability P(Y_1 = Y_2 + Y_3 | theta = 5).
Using the definition of Y_i, we can rewrite this probability as:
P(theta * X_1 = theta * X_2 + theta * X_3 | theta = 5).
Since theta = 5, we can simplify this probability to:
P(5 * X_1 = 5 * X_2 + 5 * X_3).
Dividing both sides by 5, we get:
P(X_1 = X_2 + X_3).
Since X_1, X_2, and X_3 are independent and uniformly distributed on {-1, 0, 1}, we can compute the probabilities of each individual event:
P(X_1 = 1 and X_2 = 0 and X_3 = 1) = P(X_1 = 1) * P(X_2 = 0) * P(X_3 = 1) = (1/3) * (1/3) * (1/3) = 1/27
P(X_1 = 1 and X_2 = 1 and X_3 = -1) = P(X_1 = 1) * P(X_2 = 1) * P(X_3 = -1) = (1/3) * (1/3) * (1/3) = 1/27
P(X_1 = -1 and X_2 = 0 and X_3 = -1) = P(X_1 = -1) * P(X_2 = 0) * P(X_3 = -1) = (1/3) * (1/3) * (1/3) = 1/27
P(X_1 = -1 and X_2 = -1 and X_3 = 1) = P(X_1 = -1) * P(X_2 = -1) * P(X_3 = 1) = (1/3) * (1/3) * (1/3) = 1/27
Adding up these probabilities, we get:
P(X_1 = X_2 + X_3) = 1/27 + 1/27 + 1/27 + 1/27 = 4/27
Therefore, the probability P(Y_1 = Y_2 + Y_3 | theta = 5) is equal to 4/27.