Given that $X_i \sim \text{Normal}(0, i^2)$ for $i = 1, 2, \ldots, 9$ and that $X_1, X_2, \ldots, X_9$ are independent, we need to find the variance of the random variable $Y$ defined as:

$Y = \sum_{i=1}^9 iX_i$

Steps to Find the Variance of $Y$:

1. Expression for $Y$: $Y = \sum_{i=1}^9 iX_i$

2. Variance of $Y$: Since $X_i$ are independent random variables, the variance of $Y$ is the sum of the variances of each term $iX_i$. Thus: $\text{Var}(Y) = \text{Var}\left(\sum_{i=1}^9 iX_i \right) = \sum_{i=1}^9 \text{Var}(iX_i)$

3. Variance of $iX_i$: For a constant $i$ and a random variable $X_i$: $\text{Var}(iX_i) = i^2 \text{Var}(X_i)$

4. Variance of $X_i$: Given $X_i \sim \text{Normal}(0, i^2)$, the variance of $X_i$ is $i^2$. Therefore: $\text{Var}(iX_i) = i^2 \times i^2 = i^4$

5. Summing the Variances: Now, sum the variances $\text{Var}(iX_i) = i^4$ for $i = 1$ to $9$: $\text{Var}(Y) = \sum_{i=1}^9 i^4$

6. Calculating the Sum: Calculate $\sum_{i=1}^9 i^4$: $\sum_{i=1}^9 i^4 = 1^4 + 2^4 + 3^4 + 4^4 + 5^4 + 6^4 + 7^4 + 8^4 + 9^4$ $= 1 + 16 + 81 + 256 + 625 + 1296 + 2401 + 4096 + 6561$ $= 13333$

Conclusion:

The variance of $Y$ is: $\text{Var}(Y) = 13333$

Would you like more details or have any questions about this solution? Here are some other questions you might want to explore:

1. How do you compute the mean and variance for linear combinations of random variables?
2. What is the difference between independent and identically distributed (i.i.d.) random variables?
3. How can you apply the properties of normal distributions to solve problems?
4. What are the moment generating functions and how are they used?
5. How do you generalize this result to other types of distributions?

Tip: When working with the variances of linear combinations of random variables, always remember to consider the independence of the variables, as this allows the variances to be summed directly.