Question

A random variable Y has the Density Function

f(y) = { ey, y , 0 0, elsewhere

a. Find E(e3Y/2).
b. Find the Moment Generating Function for Y.
c. Find the V(Y).

Answer 1

a. E(e^(3Y/2)) = 8. b. MGF for Y = 1 / (1 - t)^2. c. V(Y) = 1.

Step-by-step explanation:

For part a, to find E(e^(3Y/2)), use the definition of the expected value for a continuous random variable: E(g(Y)) = ∫ g(y) * f(y) dy from -∞ to ∞. Substitute g(Y) = e^(3Y/2) into this formula. The integral becomes ∫ e^(3y/2) * e^y * y dy from 0 to ∞, which evaluates to 8.

For part b, the Moment Generating Function (MGF) for a random variable Y is given by M(t) = E(e^(tY)). Substituting the given density function, the MGF becomes M(t) = ∫ e^(ty) * e^y * y dy from 0 to ∞. Simplifying this integral gives M(t) = 1 / (1 - t)^2.

For part c, the variance of Y (V(Y)) can be found using the formula V(Y) = E(Y^2) - [E(Y)]^2. First, calculate E(Y^2) by finding ∫ y^2 * e^y * y dy from 0 to ∞, which equals 2. Then, since E(Y) = 1, V(Y) = 2 - 1 = 1.

Step-by-step explanation:

Finding the expected value of e^(3Y/2) involves integrating the function e^(3y/2) * e^y * y with respect to y from 0 to ∞. Simplifying this integral results in the final answer of 8. Moving to the Moment Generating Function (MGF), substituting e^(ty) into the MGF formula yields the integral ∫ e^(ty) * e^y * y dy from 0 to ∞, which simplifies to 1 / (1 - t)^2. Lastly, calculating the variance V(Y) involves determining E(Y^2) and squaring the expected value of Y to find the variance, resulting in V(Y) = 1.

Answer 2

a.
`E(e3Y/2)` is
`-\frac25`

b. Moment Generating Function for Y is
`\[= (1)/(t+1) \cdot \infty \quad \text{(if t < -1)}\]`

c. V(Y) is
`\infty - 2^2 = \infty - 4 = \infty\]`

a. Find
`\(E\left(e^(3Y/2)\right)\)`:

The expectation of a function of a random variable can be found using the formula:

`\[E(g(Y)) = \int_(-\infty)^(\infty) g(y) f_Y(y) \, dy\]`

In this case,
`\(g(y) = e^(3y/2)\)` and
`\(f_Y(y) = e^y\)` for
`\(0 < y < \infty\), and \(f_Y(y) = 0\)` elsewhere.

So,

`\[E\left(e^(3Y/2)\right) = \int_(0)^(\infty) e^(3y/2) \cdot e^y \, dy\]`

`\[= \int_(0)^(\infty) e^((3y/2 + y)) \, dy\]`

`\[= \int_(0)^(\infty) e^((5y/2)) \, dy\]`

Now, applying the integral, we get:

`\[E\left(e^(3Y/2)\right) = \left[(2)/(5) e^((5y/2))\right]_0^\infty\]`

`\[= (2)/(5) \left(e^((5 \cdot \infty / 2)) - e^((5 \cdot 0 / 2))\right)\]`

`\[= (2)/(5) \left(0 - 1\right)\]`

`\[= -(2)/(5)\]`

b. Find the Moment Generating Function for Y.

The moment generating function (MGF) for a random variable Y is defined as:

`\[M_Y(t) = E(e^(tY))\]`

So, in this case:

`\[M_Y(t) = E(e^(tY)) = \int_(0)^(\infty) e^(ty) \cdot e^y \, dy\]`

`\[= \int_(0)^(\infty) e^((t+1)y) \, dy\]`

Now, solving the integral:

`\[M_Y(t) = \left[(1)/(t+1) e^((t+1)y)\right]_0^\infty\]`

`\[= (1)/(t+1) \left(e^((t+1) \cdot \infty) - e^((t+1) \cdot 0)\right)\]`

`\[= (1)/(t+1) \left(\infty - 1\right)\]`

`\[= (1)/(t+1) \cdot \infty \quad \text{(if t < -1)}\]`

c. Find
`\(V(Y)\)`.

The variance of a random variable Y is given by:

`\[V(Y) = E(Y^2) - [E(Y)]^2\]`

To find
`\(E(Y^2)\)`:

`\[E(Y^2) = \int_(0)^(\infty) y^2 \cdot e^y \, dy\]`

`\[= \int_(0)^(\infty) y^2 \cdot e^y \, dy\]`

Using integration by parts:

`\[E(Y^2) = \left[y^2 \cdot e^y - \int 2y \cdot e^y \, dy\right]_0^\infty\]`

`\[= \left[y^2 \cdot e^y - 2\left(y \cdot e^y - \int e^y \, dy\right)\right]_0^\infty\]`

`\[= \left[y^2 \cdot e^y - 2y \cdot e^y + 2e^y\right]_0^\infty\]`

`\[= \lim_(y \to \infty) (y^2 e^y - 2y e^y + 2e^y) - (0^2 e^0 - 2 \cdot 0 \cdot e^0 + 2e^0)\]`

`\[= \infty - 0 = \infty\]`

Since
`\(E(Y) = 2\)` (previously calculated), now we can find
`\(V(Y)\)`:

`\[V(Y) = E(Y^2) - [E(Y)]^2\]`

`\[V(Y) = \infty - 2^2 = \infty - 4 = \infty\]`

The variance of Y is infinite, indicating high variability or dispersion of Y around its mean.

The complete question is here:

A random variable Y has the Density Function

`$f(y)=\left\{e^y, y, 00\right.$`, elsewhere

a. Find
`$E\left(e^(3 Y / 2)\right)$`.

b. Find the Moment Generating Function for Y.

c. Find the
`$V(Y)$`.