Properties of Sum and Ratio of i.i.d. Exponential RVs

Let $X$ and $Y$ be independent and identically distributed (i.i.d.) exponential random variables with parameter $\lambda=1$. Their probability density function (PDF) is $f(x) = e^{-x}$ for $x > 0$. We define $W = X + Y$ and $U = X / (X + Y)$. We need to determine which of the given statements about $W$ and $U$ are true.

Verifying E(U) = 1/2

Since $X$ and $Y$ are i.i.d., the random variables $U = X/(X+Y)$ and $V = Y/(X+Y)$ have the same distribution.

Note that $U + V = X/(X+Y) + Y/(X+Y) = (X+Y)/(X+Y) = 1$.

Taking the expectation, we get $E(U + V) = E(1)$, which implies $E(U) + E(V) = 1$.

Since $U$ and $V$ have the same distribution, $E(U) = E(V)$. Therefore, $E(U) + E(U) = 1$, leading to $2E(U) = 1$, so $E(U) = 1/2$.

Statement 1 ($E(U) = 1/2$) is true.

Verifying U is Uniform on (0, 1)

To find the distribution of $U$, we can compute its cumulative distribution function (CDF), $F_U(u) = P(U \le u)$.

For $0 < u < 1$, $F_U(u) = P(X / (X+Y) \le u) = P(X \le u(X+Y)) = P(X(1-u) \le uY)$.

Since $X$ and $Y$ are i.i.d. with PDF $f(x,y) = e^{-x}e^{-y}$ for $x,y>0$, we calculate the probability:

$P(X(1-u) \le uY) = \int_0^\infty \int_0^{y \cdot u/(1-u)} e^{-x} e^{-y} dx dy$

= $\int_0^\infty \left[ -e^{-x} \right]_0^{y \cdot u/(1-u)} e^{-y} dy$

= $\int_0^\infty (1 - e^{-y \cdot u/(1-u)}) e^{-y} dy$

= $\int_0^\infty e^{-y} dy - \int_0^\infty e^{-y(1 + u/(1-u))} dy$

= $1 - \int_0^\infty e^{-y/(1-u)} dy$

= $1 - \left[ -(1-u) e^{-y/(1-u)} \right]_0^\infty$

= $1 - (0 - (-(1-u))) = 1 - (1-u) = u$.

The CDF $F_U(u) = u$ for $0 < u < 1$. This is the CDF of a Uniform(0, 1) distribution.

Statement 2 ($U$ is uniform on $(0, 1)$) is true.

Verifying W, U are Independent

We use the transformation method. Let $X = UW$ and $Y = W(1-U)$. The inverse transformation is $W = X+Y$ and $U = X/(X+Y)$.

The Jacobian of this transformation is $J = \det \begin{pmatrix} u & w \\ 1-u & -w \end{pmatrix} = u(-w) - w(1-u) = -uw - w + uw = -w$. The absolute value is $|J| = w$.

The joint PDF of $X, Y$ is $f_{X,Y}(x,y) = e^{-x} e^{-y} = e^{-(x+y)}$ for $x,y>0$. The domain corresponds to $w > 0$ and $0 < u < 1$.

The joint PDF of $W, U$ is $f_{W,U}(w, u) = f_{X,Y}(uw, w(1-u)) |J| = e^{-w} \cdot w = w e^{-w}$ for $w > 0$ and $0 < u < 1$.

Now, find the marginal PDFs:

• $f_W(w) = \int_0^1 f_{W,U}(w, u) du = \int_0^1 w e^{-w} du = w e^{-w} [u]_0^1 = w e^{-w}$ for $w > 0$.
• $f_U(u) = \int_0^\infty f_{W,U}(w, u) dw = \int_0^\infty w e^{-w} dw = \Gamma(2) = 1! = 1$ for $0 < u < 1$.

Since $f_{W,U}(w, u) = w e^{-w}$ and $f_W(w) f_U(u) = (w e^{-w}) \times 1 = w e^{-w}$, we have $f_{W,U}(w, u) = f_W(w) f_U(u)$.

Thus, $W$ and $U$ are independent.

Statement 3 ($W, U$ are independent) is true.

Verifying W, U are Uncorrelated but Dependent

As established above, $W$ and $U$ are independent.

Independence implies that the variables are uncorrelated (i.e., $Cov(W, U) = 0$).

However, independence also implies they are *not* dependent in the probabilistic sense. Since they are independent, they cannot be dependent.

Statement 4 is false.

Conclusion on True Statements

The following statements were verified as true:

• $E(U) = 1/2$
• $U$ is uniform on $(0, 1)$
• $W, U$ are independent

These correspond to the first three options listed in the question.