Let U1,U2,... be iid U(0,1). Fix d and create the d-dimensional vector U , (U1,...,Ud). Let ∥U∥ , qU12 + · · · + Ud2 be the usual Euclidean norm in d-dimensions.
(a) Fix d = 2. Calculate P(∥U∥ < 1/3). Give a numerical answer. [Hint: Draw a picture and reason geometrically. Avoid calculus.]

To solve the problem geometrically, we need to calculate the probability \( P(\| U \| < 1/3) \) for \( d = 2 \), where \( U = (U_1, U_2) \) is a two-dimensional vector with components \( U_1 \) and \( U_2 \), which are independent and identically distributed (i.i.d.) \( U(0,1) \) (uniformly distributed on the interval [0,1]).

### Step 1: Interpret the problem geometrically
For \( d = 2 \), we are working in the unit square \( [0, 1] \times [0, 1] \). The vector \( U = (U_1, U_2) \) lies within this square. The Euclidean norm of \( U \) is given by:
\[
\| U \| = \sqrt{U_1^2 + U_2^2}
\]
We are interested in finding the probability that the Euclidean norm \( \| U \| \) is less than \( \frac{1}{3} \), i.e., the probability that the point \( (U_1, U_2) \) lies within a circle of radius \( \frac{1}{3} \) centered at the origin (0,0).

### Step 2: Visualize the geometry
We are trying to find the area of the region inside the circle of radius \( \frac{1}{3} \) centered at the origin that also lies within the unit square.

- The equation of the circle is:
\[
U_1^2 + U_2^2 = \left( \frac{1}{3} \right)^2
\]
- The area of the circle with radius \( \frac{1}{3} \) is:
\[
\text{Area of full circle} = \pi \left( \frac{1}{3} \right)^2 = \frac{\pi}{9}
\]
However, only part of this circle is inside the unit square, as the circle extends outside the square in certain regions.

### Step 3: Determine the part of the circle inside the unit square
The circle is centered at \( (0, 0) \), but the square is confined to \( [0, 1] \times [0, 1] \). Hence, only the part of the circle in the first quadrant is relevant. The region of interest is a quarter of the total circle's area.

Thus, the area of the circle inside the square is:
\[
\text{Area of interest} = \frac{1}{4} \times \frac{\pi}{9} = \frac{\pi}{36}
\]

### Step 4: Calculate the probability
The probability \( P(\| U \| < 1/3) \) is the ratio of the area of the quarter-circle to the total area of the unit square. The area of the unit square is 1, so the probability is:
\[
P(\| U \| < 1/3) = \frac{\frac{\pi}{36}}{1} = \frac{\pi}{36}
\]

### Step 5: Numerical answer
Using the approximation \( \pi \approx 3.1416 \), we calculate:
\[
P(\| U \| < 1/3) \approx \frac{3.1416}{36} \approx 0.08727
\]

Thus, the numerical answer is approximately:
\[
P(\| U \| < 1/3) \approx 0.0873
\]

### Summary:
The probability that \( \| U \| < \frac{1}{3} \) for \( d = 2 \) is approximately \( 0.0873 \).

Would you like more details or have any questions? 

Here are some related questions:
1. How would the probability change if the radius were larger than \( 1/3 \)?
2. How would this approach generalize to higher dimensions, say \( d = 3 \)?
3. What is the impact of the square boundary on the probability as the dimension increases?
4. Could you determine \( P(\| U \| < r) \) for a different value of \( r \)?
5. How does the shape of the region change if we switch to a different norm (e.g., the \( L_1 \)-norm)?

**Tip:** For geometric probability, visualizing the problem often simplifies the computation, especially in cases where calculus can be avoided.

Let U1,U2,... be iid U(0,1). Fix d and create the d-dimensional vector U , (U1,...,Ud). Let ∥U∥ , sqrt(U1^2 + ... + Ud^2) be the usual Euclidean norm in d-dimensions. (a) Fix d = 2. Calculate P(∥U∥ < 1/3). Give a numerical answer. [Hint: Draw a picture and reason geometrically. Avoid calculus.]

This problem explores geometric probability using a uniform distribution in two dimensions. Learn how to calculate the probability that the Euclidean norm of a vector ∥U∥ is less than 1/3, including step-by-step reasoning and numerical approximation. Suitable for undergraduate students.

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