Manhole explosions (usually caused by gas leaks and sparks) are on the rise in your city. On any given day, the manhole cover near your house explodes with some unknown probability, which is the same across all days. We model this unknown probability of explosion as a random variable Q, which is uniformly distributed between 0 and 0.1. Let Xi be a Bernoulli random variable that indicates whether the manhole cover near your house explodes on day i (where today is day 1).

Question

Manhole explosions (usually caused by gas leaks and sparks) are on the rise in your city. On any given day, the manhole cover near your house explodes with some unknown probability, which is the same across all days. We model this unknown probability of explosion as a random variable Q, which is uniformly distributed between 0 and 0.1. Let Xi be a Bernoulli random variable that indicates whether the manhole cover near your house explodes on day i (where today is day 1).

Give numerical answers for parts (1) and (2).

E[Xi]=

- unanswered

var(Xi)=

- unanswered

Let A be the event that the manhole cover did not explode yesterday (i.e., X0=0). Find the conditional PDF of Q given A. Express your answer in terms of q using standard notation .

For 0≤q≤0.1, fQ∣A(q)=

Answer 1

To find the numerical answers for parts (1) and (2), we need to understand the concept of expected value (E[Xi]) and variance (var(Xi)).

1) The expected value (E[Xi]) of a random variable measures the average value we would expect to observe if we repeatedly sample from that random variable. In this case, Xi represents whether the manhole cover near your house explodes on day i. If we assume that the probability of explosion Q is uniformly distributed between 0 and 0.1, then the expected value E[Xi] can be calculated as:

E[Xi] = P(Xi = 1) * 1 + P(Xi = 0) * 0

Since Xi is a Bernoulli random variable, it can only take values 0 or 1. The probability of Xi = 1 is equal to the probability of explosion, which is Q. The probability of Xi = 0 is equal to the probability of no explosion, which is 1 - Q. Therefore, we can calculate E[Xi] as:

E[Xi] = Q * 1 + (1 - Q) * 0
= Q

So the answer for part (1) is E[Xi] = Q.

2) The variance (var(Xi)) of a random variable measures how spread out the observed values are around the expected value. It quantifies the variability or dispersion in the data. For a Bernoulli random variable, the variance can be calculated using the formula:

var(Xi) = E[Xi^2] - E[Xi]^2

The expected value E[Xi] is already determined to be Q from part (1). To calculate E[Xi^2], we need to consider the probability distribution of the random variable Xi^2:

P(Xi^2 = 1) = P(Xi = 1) * 1 + P(Xi = 0) * 0
= Q * 1 + (1 - Q) * 0
= Q

P(Xi^2 = 0) = P(Xi = 1) * 0 + P(Xi = 0) * 1
= Q * 0 + (1 - Q) * 1
= 1 - Q

Now we can calculate E[Xi^2] as:

E[Xi^2] = P(Xi^2 = 1) * 1 + P(Xi^2 = 0) * 0
= Q * 1 + (1 - Q) * 0
= Q

Substituting these values into the variance formula, we get:

var(Xi) = E[Xi^2] - E[Xi]^2
= Q - Q^2
= Q(1 - Q)

So the answer for part (2) is var(Xi) = Q(1 - Q).

Moving onto the next part of the question:

Let A be the event that the manhole cover did not explode yesterday, i.e., X0 = 0. We need to find the conditional PDF of Q given A, denoted as fQ|A(q).

Since A has occurred, we know that X0 = 0. This means that the probability of explosion on day 0 is 0. Therefore, the conditional PDF of Q given A can be expressed as follows:

fQ|A(q) = {1/0.1 if 0 ≤ q ≤ 0.1,
0 otherwise}