Problem Statement
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 \(\frac{\mathbb{E}[Y]}{Var(Y)}\)
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Solution
Step 1 - Compute Conditional Expectation \(\mathbb{E}[Y \mid X]\) and Conditional Variance \(Var(Y \mid X)\)
For \(Y\) uniform on \((X,1)\), the conditional expectation is:
\[\mathbb{E}[Y \mid X] = \frac{X+1}{2}\]How? Its obvious that its gonna be the middle point! Alternatively, you can integrate and compute using the standard method.
Similarly, for Conditional Variance \(Var[Y \mid X]\), we have
\[Var[Y \mid X] = \frac{(1-X)^2}{12}\]Step 2 - Total Expectation
Using the law of total expectation:
\[\mathbb{E}[Y] = \mathbb{E}[\mathbb{E}[Y \mid X]] = \mathbb{E}\left[\frac{X+1}{2}\right] = \int_0^1 \frac{x+1}{2} dx = \left[\frac{x^2}{4} + \frac{x}{2}\right]_0^1 = \frac{1}{4} + \frac{1}{2} = \frac{3}{4}\]Step 3 - Computing Total Variance
Using the law of total variance:
\[Var(Y) = \mathbb{E}[Var(Y \mid X)] + Var(\mathbb{E}[Y \mid X])\]First, compute \(\mathbb{E}[Var(Y \mid X)]\):
\[\mathbb{E}[Var(Y \mid X)] = \int_0^1 \frac{(1-x)^2}{12} dx\] \[= \frac{1}{12} \int_0^1 (1-x)^2 dx = \frac{1}{12} \left[\frac{-1}{3}(1-x)^3\right]_0^1 = \frac{1}{12} \cdot \frac{1}{3} = \frac{1}{36}\]Next, compute \(Var(\mathbb{E}[Y \mid X])\):
\[Var(\mathbb{E}[Y \mid X]) = \mathbb{E}\left[\left(\frac{X+1}{2}\right)^2\right] - \left(\mathbb{E}\left[\frac{X+1}{2}\right]\right)^2\] \[= \int_0^1 \frac{(x+1)^2}{4} dx - \left(\frac{3}{4}\right)^2 = \frac{1}{4} \int_0^1 (x^2 + 2x + 1) dx - \frac{9}{16}\] \[= \frac{1}{4} \left[\frac{x^3}{3} + x^2 + x\right]_0^1 - \frac{9}{16} = \frac{1}{4} \left(\frac{1}{3} + 1 + 1\right) - \frac{9}{16} = \frac{1}{4} \cdot \frac{7}{3} - \frac{9}{16} = \frac{7}{12} - \frac{9}{16} = \frac{1}{48}\]Therefore:
\[Var(Y) = \frac{1}{36} + \frac{1}{48} = \frac{7}{144}\]Final Answer
\[\frac{\mathbb{E}[Y]}{Var(Y)} = \frac{3/4}{7/144} = \frac{108}{7}\]💡 Did you enjoy this problem? Check out more puzzles in the Problems section!