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Math  /  Data & Statistics

Question1. (5.2 MLE) A random sample of size n=9n=9 is taken from the probability function: pY(k;λ)=λk(1−λ)1−k,k=0,1;0<λ<1p_{Y}(k ; \lambda)=\lambda^{k}(1-\lambda)^{1-k}, k=0,1 ; \quad 0<\lambda<1
The samples are Y1=1,Y2=1,Y3=1,Y4=0,Y5=0,Y6=1,Y7=1,Y8=0,Y9=1Y_{1}=1, Y_{2}=1, Y_{3}=1, Y_{4}=0, Y_{5}=0, Y_{6}=1, Y_{7}=1, Y_{8}=0, Y_{9}=1. Find the maximum likelihood estimate λ^\hat{\lambda} for λ\lambda.

Studdy Solution

STEP 1

What is this asking? We're trying to find the best guess for a hidden probability (λ\lambda) based on some observations of ones and zeros. Watch out! Don't mix up the number of ones and zeros with the *probability* of getting a one or a zero.

STEP 2

1. Define the likelihood function
2. Take the logarithm of the likelihood function
3. Optimize the log-likelihood

STEP 3

Alright, let's **define the likelihood function**, which tells us how likely our observations are given a specific value of λ\lambda.
Think of it like this: we've flipped a weird coin 9 times, and we got some ones (heads) and some zeros (tails).
We want to figure out the probability (λ\lambda) of getting a one (heads) that best explains our results!

STEP 4

The probability function for a single observation YiY_i is given by pY(k;λ)=λk(1−λ)1−kp_{Y}(k ; \lambda)=\lambda^{k}(1-\lambda)^{1-k}.
Since our samples are independent, the likelihood function is the product of the individual probabilities for each observation:
L(λ)=∏i=19λYi(1−λ)1−Yi L(\lambda) = \prod_{i=1}^{9} \lambda^{Y_i}(1-\lambda)^{1-Y_i}

STEP 5

Now, products can be tricky to work with, so let's take the **natural logarithm** of the likelihood function.
This turns the product into a sum, which is *way* easier to handle, and it doesn't change where the maximum occurs!
This gives us the **log-likelihood**:
ln⁡L(λ)=∑i=19ln⁡(λYi(1−λ)1−Yi)=∑i=19[Yiln⁡(λ)+(1−Yi)ln⁡(1−λ)] \ln L(\lambda) = \sum_{i=1}^{9} \ln(\lambda^{Y_i}(1-\lambda)^{1-Y_i}) = \sum_{i=1}^{9} [Y_i \ln(\lambda) + (1-Y_i) \ln(1-\lambda)]

STEP 6

Let's simplify this further.
We know the values of our YiY_i’s.
We have six ones and three zeros.
So, the sum becomes:
ln⁡L(λ)=6ln⁡(λ)+3ln⁡(1−λ) \ln L(\lambda) = 6 \ln(\lambda) + 3 \ln(1-\lambda)

STEP 7

To find the **maximum likelihood estimate**, we need to find the value of λ\lambda that maximizes our log-likelihood.
We do this by taking the derivative with respect to λ\lambda and setting it equal to zero.
ddλln⁡L(λ)=6λ−31−λ \frac{d}{d\lambda} \ln L(\lambda) = \frac{6}{\lambda} - \frac{3}{1-\lambda}

STEP 8

Setting the derivative equal to zero and solving for λ\lambda:
6λ−31−λ=0 \frac{6}{\lambda} - \frac{3}{1-\lambda} = 0 6λ=31−λ \frac{6}{\lambda} = \frac{3}{1-\lambda} 6(1−λ)=3λ 6(1-\lambda) = 3\lambda 6−6λ=3λ 6 - 6\lambda = 3\lambda 6=9λ 6 = 9\lambda λ^=69=23 \hat{\lambda} = \frac{6}{9} = \frac{2}{3}

STEP 9

Our **maximum likelihood estimate** for λ\lambda is λ^=23\hat{\lambda} = \frac{2}{3}.
So, based on our observations, the best guess for the probability of getting a one is 23\frac{2}{3}!

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