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velikii [3]
2 years ago
13

Let X be a Bernoulli rv with pmf as in Example 3.18. a. Compute E(X2 ). b. Show that V(X) 5 p(1 2 p). c. Compute E(X79).

Mathematics
1 answer:
spayn [35]2 years ago
5 0

The Bernoulli distribution is a distribution whose random variable can  only take 0 or 1

  • The value of E(x2) is p
  • The value of V(x) is p(1 - p)
  • The value of E(x79) is p

<h3>How to compute E(x2)</h3>

The distribution is given as:

p(0) = 1 - p

p(1) = p

The expected value of x2, E(x2) is calculated as:

E(x^2) = \sum x^2 * P(x)

So, we have:

E(x^2) = 0^2 * (1- p) + 1^2 * p

Evaluate the exponents

E(x^2) = 0 * (1- p) + 1 * p

Multiply

E(x^2) = 0 +p

Add

E(x^2) = p

Hence, the value of E(x2) is p

<h3>How to compute V(x)</h3>

This is calculated as:

V(x) = E(x^2) - (E(x))^2

Start by calculating E(x) using:

E(x) = \sum x * P(x)

So, we have:

E(x) = 0 * (1- p) + 1 * p

E(x) = p

Recall that:

V(x) = E(x^2) - (E(x))^2

So, we have:

V(x) = p - p^2

Factor out p

V(x) = p(1 - p)

Hence, the value of V(x) is p(1 - p)

<h3>How to compute E(x79)</h3>

The expected value of x79, E(x79) is calculated as:

E(x^{79}) = \sum x^{79} * P(x)

So, we have:

E(x^{79}) = 0^{79} * (1- p) + 1^{79} * p

Evaluate the exponents

E(x^{79}) = 0 * (1- p) + 1 * p

Multiply

E(x^{79}) = 0 + p

Add

E(x^{79}) = p

Hence, the value of E(x79) is p

Read more about probability distribution at:

brainly.com/question/15246027

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