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).
Show 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.
Step 1: Set the target formula.
We want $\mathrm{Var}(X)+\mathrm{Var}(Y)$. Using $\mathrm{Var}(X)=E(X^2)-E(X)^2$ and the same for $Y$, we need four numbers: $E(X)$, $E(Y)$, $E(X^2)$, $E(Y^2)$.
Step 2: Compute E(X) as a direct double integral.
\[ E(X)=\int_0^1\int_0^1 x(x+y)\,dy\,dx=\int_0^1\Big(x^2+\frac{x}{2}\Big)dx=\frac13+\frac14=\frac{7}{12}. \]
Step 3: Compute E(Y) the same way.
By swapping the roles of $x$ and $y$ in the integrand $x+y$ (which is symmetric), the same computation gives
\[ E(Y)=\int_0^1\int_0^1 y(x+y)\,dx\,dy=\frac{7}{12}. \]
So $E(X)=E(Y)$ without needing a separate calculation.
Step 4: Compute E(X^2) as a direct double integral.
\[ E(X^2)=\int_0^1\int_0^1 x^2(x+y)\,dy\,dx=\int_0^1\Big(x^3+\frac{x^2}{2}\Big)dx=\frac14+\frac16=\frac{5}{12}. \]
Step 5: Get E(Y^2) by the same symmetry shortcut.
Swapping $x$ and $y$ again in the symmetric integrand gives $E(Y^2)=\frac{5}{12}$ too, so both squared moments match.
Step 6: Combine into the variance sum.
\[ \mathrm{Var}(X)+\mathrm{Var}(Y)=\Big[E(X^2)-E(X)^2\Big]+\Big[E(Y^2)-E(Y)^2\Big]=2\Big(\frac{5}{12}-\frac{49}{144}\Big)=2\cdot\frac{11}{144}=\frac{11}{72}. \]
Step 7: Scale by 72.
\[ 72\cdot\frac{11}{72}=11. \]
Final Answer:
Working straight from the double integral, without stopping to write out the marginal density formula, gives the same clean value.
\[ \boxed{11} \]
