Question

Let \(X\) and \(Y\) be two continuous random variables having the following joint probability density function
\[ f(x,y)=\begin{cases} x+y & \text{if } 0<x<1,\ 0<y<1 \\ 0 & \text{otherwise}. \end{cases} \]
Then \(72\big(\mathrm{Var}(X)+\mathrm{Var}(Y)\big)\) equals ________ (answer in integer).

Hint:

Hint:
Find the marginal density of X by integrating out y, then compute \(E(X)\) and \(E(X^2)\) to get \(\mathrm{Var}(X)\). Use the symmetry of \(f(x,y)=x+y\) to get \(\mathrm{Var}(Y)\) the same way.

Answer 1

Correct Answer: 11

Step 1: Check that f is a valid joint density.
Before using \(f(x,y)=x+y\) on the unit square, note that it is nonnegative there, and
\[ \int_0^1\int_0^1 (x+y)\,dy\,dx=\int_0^1\Big(x+\tfrac12\Big)dx=\tfrac12+\tfrac12=1, \]
so it is a proper density.

Step 2: Find the marginal density of X.
Integrate out y:
\[ f_X(x)=\int_0^1 (x+y)\,dy=x\cdot 1+\Big[\frac{y^2}{2}\Big]_0^1=x+\frac{1}{2}, \qquad 0<x<1. \]

Step 3: Find E(X).
\[ E(X)=\int_0^1 x\Big(x+\frac12\Big)dx=\int_0^1 x^2\,dx+\frac12\int_0^1 x\,dx=\frac13+\frac12\cdot\frac12=\frac13+\frac14=\frac{7}{12}. \]

Step 4: Find E(X^2).
\[ E(X^2)=\int_0^1 x^2\Big(x+\frac12\Big)dx=\int_0^1 x^3\,dx+\frac12\int_0^1 x^2\,dx=\frac14+\frac12\cdot\frac13=\frac14+\frac16=\frac{5}{12}. \]

Step 5: Find Var(X).
\[ \mathrm{Var}(X)=E(X^2)-[E(X)]^2=\frac{5}{12}-\Big(\frac{7}{12}\Big)^2=\frac{5}{12}-\frac{49}{144}=\frac{60}{144}-\frac{49}{144}=\frac{11}{144}. \]

Step 6: Use symmetry for Y.
The density \(f(x,y)=x+y\) is symmetric in x and y, so the marginal of Y has exactly the same shape, \(f_Y(y)=y+1/2\). By the identical calculation,
\[ \mathrm{Var}(Y)=\frac{11}{144}. \]

Step 7: Add the variances and scale by 72.
\[ \mathrm{Var}(X)+\mathrm{Var}(Y)=\frac{11}{144}+\frac{11}{144}=\frac{22}{144}=\frac{11}{72}. \]
\[ 72\big(\mathrm{Var}(X)+\mathrm{Var}(Y)\big)=72\cdot\frac{11}{72}=11. \]

Final Answer:
The scaled sum of variances comes out to a clean integer. \[ \boxed{11} \]