Question

Let \( X \) be a random variable with the moment generating function \[ M_X(t) = \left( \frac{e^{t/2} + e^{-t/2}}{2} \right)^2, \quad -\infty<t<\infty. \] Using Chebyshev's inequality, the upper bound for \( P \left( |X|>\frac{2}{\sqrt{3}} \right) \) equals ...............

Hint:

Chebyshev's inequality gives an upper bound on the probability that a random variable deviates from its mean by more than \( k \) standard deviations. It is a useful tool for bounding probabilities without knowing the exact distribution of the random variable.

Answer 1

Correct Answer: 0.75

Step 1: Use the moment generating function.
The moment generating function (MGF) for \( X \) is given by: \[ M_X(t) = \left( \frac{e^{t/2} + e^{-t/2}}{2} \right)^2 = \cosh^2\left(\frac{t}{2}\right). \] Step 2: Apply Chebyshev’s inequality.
Chebyshev’s inequality states that for any random variable with mean \( \mu \) and variance \( \sigma^2 \), \[ P(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}. \] Here, the mean \( \mu = 0 \) and variance \( \sigma^2 = 1 \), so the inequality becomes: \[ P(|X| \geq k) \leq \frac{1}{k^2}. \] We want to find \( P \left( |X|>\frac{2}{\sqrt{3}} \right) \), so we set \( k = \frac{2}{\sqrt{3}} \).
Step 3: Calculate the probability.
Using Chebyshev’s inequality, we get: \[ P \left( |X|>\frac{2}{\sqrt{3}} \right) \leq \frac{1}{\left(\frac{2}{\sqrt{3}}\right)^2} = \frac{1}{\frac{4}{3}} = \frac{3}{4}. \]
Final Answer: \[ \boxed{\frac{3}{4}}. \]