Final answer:
The expected sum of a series of random variables Y1, Y2, ..., YX dependent on random variable X, with X following a normal distribution with mean 10 and variance 500 and Y uniformly distributed on [0, 1], is the product of the expected values of X and Y, resulting in 5.
Step-by-step explanation:
Expected Sum of Random Variables
The question requires us to understand the expected sum of a series of random variables Y1, Y2, ..., YX dependent on another random variable X, with the specific distributions provided for both X and Y variables.
a. The probability density function (PDF) of the random variable X is that of a normal distribution. Given that X has a mean (μ) of 10 and variance (σ^2) of 500, the PDF is defined mathematically as f(x) = (1/(σ√(2π))) * e^((-1/2)((x-μ)/σ)^2), where σ is the standard deviation.
b. The expected value of X is simply the mean of the normal distribution, which is 10.
c. The probability distribution of Y is uniform on the interval [0, 1], which means that any value within this range is equally likely. Its PDF is f(y) = 1 for 0 ≤ y ≤ 1 and f(y) = 0 otherwise.
d. To evaluate the expected sum of Y1 ... YX, we calculate the expected value of a single Yi (which is 0.5 for a uniform distribution on [0,1]) and then multiply this by the expected value of X. The expected sum is therefore E[sum of Y's] = E[X] * E[Y] = 10 * 0.5 = 5.