Question

If, joint distribution function of two random variables X and Y is given by \(F_{X,Y}(x,y) = \begin{cases} 1 - e^{-x} - e^{-y} + e^{-(x+y)} & ; x>0; y>0 \\ 0 & ; \text{otherwise} \end{cases}\), then Var(\(X\)) is

• A. 1
• B. 2
• C. 0
• D. \( \frac{1}{2} \)

Hint:

When given a joint CDF, check if it can be factored into a product of functions of \(x\) and \(y\) only, i.e., \(F(x,y) = G(x)H(y)\). If it can, the variables are independent, and \(G(x)\) and \(H(y)\) are their respective marginal CDFs. Recognizing standard distributions like the exponential from their CDF (\(1-e^{-\lambda x}\)) or PDF (\(\lambda e^{-\lambda x}\)) can save a lot of time.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The problem provides the joint cumulative distribution function (CDF) of two random variables X and Y. To find the variance of X, we first need to find the marginal CDF of X, then the marginal probability density function (PDF) of X. From the PDF, we can identify the distribution or calculate the variance directly.

Step 2: Key Formula or Approach:
1. Find the marginal CDF of X, \(F_X(x)\), by taking the limit of the joint CDF as \(y \to \infty\). \[ F_X(x) = \lim_{y \to \infty} F_{X,Y}(x,y) \] 2. Find the marginal PDF of X, \(f_X(x)\), by differentiating the marginal CDF. \[ f_X(x) = \frac{d}{dx} F_X(x) \] 3. Recognize the distribution from its PDF. The PDF \(f(x) = \lambda e^{-\lambda x}\) for \(x>0\) corresponds to an exponential distribution with rate \(\lambda\). 4. The variance of an exponential distribution is given by \(\text{Var}(X) = \frac{1}{\lambda^2}\).

Step 3: Detailed Explanation:
First, we find the marginal CDF of X for \(x>0\): \[ F_X(x) = \lim_{y \to \infty} (1 - e^{-x} - e^{-y} + e^{-(x+y)}) \] As \(y \to \infty\), the term \(e^{-y} \to 0\) and the term \(e^{-(x+y)} = e^{-x}e^{-y} \to 0\). \[ F_X(x) = 1 - e^{-x} - 0 + 0 = 1 - e^{-x} \] So, the marginal CDF of X is \(F_X(x) = 1 - e^{-x}\) for \(x>0\). Next, we find the marginal PDF of X by differentiating \(F_X(x)\): \[ f_X(x) = \frac{d}{dx}(1 - e^{-x}) = -(-e^{-x}) = e^{-x} \] The PDF is \(f_X(x) = e^{-x}\) for \(x>0\). This is the PDF of an exponential distribution with rate parameter \(\lambda = 1\). Finally, we find the variance of X. For an exponential distribution with rate \(\lambda\), the variance is \(1/\lambda^2\). \[ \text{Var}(X) = \frac{1}{1^2} = 1 \] Alternatively, notice that the joint CDF can be factored: \[ F_{X,Y}(x,y) = 1 - e^{-x} - e^{-y} + e^{-x}e^{-y} = (1 - e^{-x})(1 - e^{-y}) = F_X(x)F_Y(y) \] This shows that X and Y are independent, and both follow an exponential distribution with \(\lambda = 1\).
Step 4: Final Answer:
The variance of X is 1.