Modified RNG

Hard

Expectations Conditional Expectation Conditional Variance

Problem

Jimmy picks a number uniformly at random from (0,1). If Jimmy chooses x, then Jon picks a number from (x,1) uniformly at random. If Y represents the number Jon selects, in simplest form, find $\frac{E [ Y ]}{Va r ( Y )}$

Original Problem Link: Click here

Solution

Step 1 - Compute Conditional Expectation $E [Y ∣ X]$ and Conditional Variance $Va r (Y ∣ X)$

For $Y$ uniform on $(X, 1)$ , the conditional expectation is:

$E [Y ∣ X] = \frac{X + 1}{2}$

How? Its obvious that its gonna be the middle point! Alternatively, you can integrate and compute using the standard method.

Similarly, for Conditional Variance $Va r [Y ∣ X]$ , we have

$Va r [Y ∣ X] = \frac{( 1 - X ) ^{2}}{12}$

Step 2 - Total Expectation

Using the law of total expectation:

$E [Y] = E [E [Y ∣ X]] = E [\frac{X + 1}{2}] = \int_{0}^{1} \frac{x + 1}{2} d x = [\frac{x ^{2}}{4} + \frac{x}{2}]_{0}^{1} = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}$

Step 3 - Computing Total Variance

Using the law of total variance:

$Va r (Y) = E [Va r (Y ∣ X)] + Va r (E [Y ∣ X])$

First, compute $E [Va r (Y ∣ X)]$ :

$E [Va r (Y ∣ X)] = \int_{0}^{1} \frac{( 1 - x ) ^{2}}{12} d x$

$= \frac{1}{12} \int_{0}^{1} (1 - x)^{2} d x = \frac{1}{12} [\frac{- 1}{3} (1 - x)^{3}]_{0}^{1} = \frac{1}{12} \cdot \frac{1}{3} = \frac{1}{36}$

Next, compute $Va r (E [Y ∣ X])$ :

$Va r (E [Y ∣ X]) = E [(\frac{X + 1}{2})^{2}] - (E [\frac{X + 1}{2}])^{2}$

$= \int_{0}^{1} \frac{( x + 1 ) ^{2}}{4} d x - (\frac{3}{4})^{2} = \frac{1}{4} \int_{0}^{1} (x^{2} + 2 x + 1) d x - \frac{9}{16}$

$= \frac{1}{4} [\frac{x ^{3}}{3} + x^{2} + x]_{0}^{1} - \frac{9}{16} = \frac{1}{4} (\frac{1}{3} + 1 + 1) - \frac{9}{16} = \frac{1}{4} \cdot \frac{7}{3} - \frac{9}{16} = \frac{7}{12} - \frac{9}{16} = \frac{1}{48}$

Therefore:

$Va r (Y) = \frac{1}{36} + \frac{1}{48} = \frac{7}{144}$

Final Answer

$\frac{E [ Y ]}{Va r ( Y )} = \frac{3/4}{7/144} = \frac{108}{7}$