The geometric distribution describes the total number of...

The geometric distribution describes the total number of experiments k that one must perform before getting a success, assuming that each experiment is a Bernoulli trial with probability p. If X ∼ geom(p), then P(X = k) = (1 − p)k−1p 1. (2) Compute the likelihood in the parameter θ of a random sample X1, . . . , Xn of i.i.d geom(θ) random variables. 2. (2) Compute the log-likelihood for the previous sample. 3. (2) Compute the MLE. 4. (1) A company produces light-bulbs. They must estimate the probability that a light-bulb is faulty. They test random light-bulbs until they find one faulty and they record how many were tested before finding the faulty one. They repeat this process 5 times and record the following results 4, 32, 11, 22, 21 Estimate the probability θ that a light-bulb is faulty.

Transcript

00:01 So in this question, the first part asks us to compute the likelihood of the parameter theta here.

00:08 So theta given the sample that we have.

00:12 So basically the idea here is that the likelihood is the density of our sample given the parameter theta.

00:23 Considering that for one observation here, x1, for example, we have that x1, has the x1, have a distribution given by the geometric distribution which is given 1 minus theta in this case x1 minus 1 times theta and here we are going to use that these observations are independent which means that and also identically distributed so they have all the same parameter so basically the idea he is that that we can first open this as the product of all the observations from one to n.

01:14 And here we can put the geometric distribution, which is the same because all of them has the same parameter.

01:25 So basically, the idea here is that i'm going to put the product here, and then i'm going to repeat this here, because it will be the same for all the observations, since they are identically distributed.

01:39 So basically what we need to do is this and this.

01:43 Now what we need to check is this one here when we multiply or when we open this product.

01:50 Basically what we're going to have is 1 minus theta x1 minus 1 times 1 minus theta x2 minus 1 until we get 1.

02:01 Minus theta x n minus 1.

02:06 And of course, this one here does not depend on i.

02:09 So basically since we are multiplying this n times, this means that this will be theta at power n.

02:18 And because we have the same value here, this means that we should sum all the exponents.

02:26 So basically what we're going to have is 1 minus theta, the sum of x i minus n and we need to repeat theta m so this is the likelihood so this is the first part that we have here to compute the second part is to compute the log likelihood so the log likelihood basically is the log of this likelihood so you just need to apply the log function here and when we apply the log function here, the exponent will like a fall down.

03:03 So we have this here minus n, then this log of 1 minus theta.

03:13 And then we have the sum because the log of a multiplication is the sum of the logs.

03:18 So basically we're going to have n log of theta.

03:22 Now, considering this, we should find the maximum likelihood estimated for theta.

03:31 So to do this, we should compute what is the derivative of our log likelihood with respect with our parameter.

03:42 So in this case, theta...

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