Question

Let \(X\) and \(Y\) be independent and identically distributed normal random variables. If
\[ P(X+2Y\le3)=P(2X-Y\ge4), \]
then \(E(X)\) is

• A. \(\dfrac{7}{3}\)
• B. \(\dfrac{7}{4}\)
• C. \(\dfrac{3}{7}\)
• D. \(\dfrac{4}{7}\)

Hint:

\(X+2Y\) and \(2X-Y\) are both normal with the same variance \(5\sigma^2\). Equal probabilities mean equal standardized cutoffs, so set \(\dfrac{3-3\mu}{\sqrt5}=\dfrac{\mu-4}{\sqrt5}\) and solve for \(\mu\).

Answer 1

The Correct Option is B

Step 1: Set up the distributions of the two linear combinations.
Let \(X\) and \(Y\) be independent and identically distributed as \(N(\mu,\sigma^2)\), where \(\mu=E(X)=E(Y)\). We need two new random variables built from \(X\) and \(Y\): \[ U=X+2Y, \qquad V=2X-Y. \] Since \(U\) and \(V\) are linear combinations of independent normal variables, both \(U\) and \(V\) are themselves normally distributed.

Step 2: Find the mean and variance of \(U=X+2Y\).
\[ E(U)=E(X)+2E(Y)=\mu+2\mu=3\mu. \] \[ Var(U)=Var(X)+4\,Var(Y)=\sigma^2+4\sigma^2=5\sigma^2. \] So \(U\sim N(3\mu,5\sigma^2)\).

Step 3: Find the mean and variance of \(V=2X-Y\).
\[ E(V)=2E(X)-E(Y)=2\mu-\mu=\mu. \] \[ Var(V)=4\,Var(X)+Var(Y)=4\sigma^2+\sigma^2=5\sigma^2. \] So \(V\sim N(\mu,5\sigma^2)\). Notice \(U\) and \(V\) share the same variance, \(5\sigma^2\).

Step 4: Write both probabilities using the standard normal.
\[ P(U\le3)=\Phi\!\left(\frac{3-3\mu}{\sigma\sqrt5}\right). \] For the second one, we first flip the inequality: \[ P(V\ge4)=1-P(V<4)=1-\Phi\!\left(\frac{4-\mu}{\sigma\sqrt5}\right)=\Phi\!\left(\frac{\mu-4}{\sigma\sqrt5}\right), \] using the symmetry property \(1-\Phi(z)=\Phi(-z)\).

Step 5: Equate the two probabilities.
The question tells us these two probabilities are equal, so \[ \Phi\!\left(\frac{3-3\mu}{\sigma\sqrt5}\right)=\Phi\!\left(\frac{\mu-4}{\sigma\sqrt5}\right). \] Since \(\Phi\) is a strictly increasing, one to one function, the two arguments must be equal: \[ \frac{3-3\mu}{\sigma\sqrt5}=\frac{\mu-4}{\sigma\sqrt5}. \]

Step 6: Solve for \(\mu\).
The common factor \(\sigma\sqrt5\) cancels from both sides, leaving \[ 3-3\mu=\mu-4 \] \[ 3+4=\mu+3\mu \] \[ 7=4\mu \] \[ \mu=\frac{7}{4}. \]

Step 7: Check the other options.
Option (A) \(7/3\), option (C) \(3/7\), and option (D) \(4/7\) all come from mixing up the coefficients while combining the two equations. Only careful bookkeeping of the coefficients \(3\) and \(4\) gives the correct value \(7/4\).

Final Answer:
Since \(E(X)=\mu\), we get \[ \boxed{E(X)=\frac{7}{4}} \]