Problem Analysis #36

[THOUGHT] To compute the variance of a random variable X, we use the formula: V(X) = E[(X - E[X])^2]

Here,

1. We first need to find E[X], the expected value (mean) of X.
2. Then we calculate E[(X - E[X])^2] by squaring each deviation from the mean and taking their expectation.

For our case, X is a discrete random variable with given probability mass function (PMF). The PMF gives us the probability of each possible value of X.

1. Calculate E[X]: E[X] = Σ(x * P(X=x)) over all x = 0*(1/12) + 1*(1/4) + 2*(3/12) + 3*(1/4) + 4*a = 1/4 + 3/4 + a

Since the problem doesn't specify a value for a, we leave it as is.

2. Calculate E[(X-E[X])^2]: We need to compute (x - E[X])^2 * P(X=x) for each x and sum them up. For x=0, 1, 2, 3: (x - E[X])^2 = (x - (1/2 + 3/2 + a))^2 = ((-1/2 - a)^2), ((1/4 - a)^2), ((5/4 - a)^2), ((3/2 - a)^2) P(X=x) = 1/12 each

For x=4: (x - E[X])^2 = (x - (1/2 + 3/2 + a))^2 = (4 - (1/2 + 3/2 + a))^2 P(X=x) = a

Now we sum these up and simplify.

V(X) = Σ((x-E[X])^2 * P(X=x)) = (-1/2 - a)^2*(1/12) + (1/4 - a)^2*(1/12) + (5/4 - a)^2*(1/12) + (3/2 - a)^2*(1/12) + (4-(1/2+3/2+a))^2*a = ... (long expression)

3. Compute V(1/a X): Given that variance is scale-invariant, i.e., V(aX) = a^2V(X), we have: V(1/a X) = (1/a)^2 * V(X) = 1/(a^2) * ...

We now use the above expression for V(X) and simplify to get the final answer.

[Python]

import sympy

# define symbols
a = sympy.Symbol('a')

# calculate E[X]
E_X = (1/4 + 3/4 + a)

# calculate terms of variance expression
term_0 = (-1/2 - a)**2 * (1/12)
term_1 = (1/4 - a)**2 * (1/12)
term_2 = (5/4 - a)**2 * (1/12)
term_3 = (3/2 - a)**2 * (1/12)
term_4 = (4-(1/2+3/2+a))**2 * a

# calculate variance V(X)
V_X = sympy.simplify(term_0 + term_1 + term_2 + term_3 + term_4)

# calculate V(1/a X)
V_1_over_a_X = (1/(a**2)) * V_X

print(V_1_over_a_X)

The output will be a complex expression in terms of 'a'. To get the final variance value, you would need to specify a particular value for 'a' and substitute it into this expression.)

[THOUGHT] Given the expression for V(1/a X) from the Python output: V_1_over_a_X = (1.0a**3 - 3.66666666666667a2 + 3.58333333333333*a + 0.34375)/a2

We can simplify this by dividing each term by a^2, which is the common denominator: V(1/a X) = 1.0*a - (3.66666666666667) + (3.58333333333333)/a + 0.34375/a**2

Now we observe that a is a nonzero constant and can be treated as such. We thus compute the values: 1.0*a = a

• (3.66666666666667)

• (3.58333333333333)/a
• 0.34375/a**2

Let's calculate these values for a few specific cases:

a=1: V(1/a X) = 1 - 3.6667 + 3.5833 + 0.3438 = 1.5606 a=10: V(1/a X) = 10 - 36.6667 + 35.8333 + 0.3438 = 9.9102

As a goes to infinity (very large values), the term 0.34375/a^2 becomes very small and can be neglected. Hence, for very large a, V(1/a X) ≈ a - 3.6667 + 3.5833 ≈ a - 0.0834

This suggests that for very large values of a, the variance of 1/a X is approximately equal to a.

\boxed{9.9102} (for a=10) \boxed{1.5606} (for a=1) (Values are rounded to four decimal places.)