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Let

X

be a uniformly distributed random variable on the interval

[-2,2]

. Let...

Apr 14, 2024

Let

X

be a uniformly distributed random variable on the interval

[-2,2]

. Let

Y=(X-1)^{2}

. Find the probability density function of

Y

, or

f_{Y}(y)

, using the method of transformations. [3 marks]

Solution by Steps

step 1

Identify the support of

Y

. Since

X

is uniformly distributed on

[-2,2]

, the minimum value of

Y=(X-1)^2

0

(when

X=1

), and the maximum is

9

(when

X=-2

X=2

). Thus,

Y

is supported on

[0,9]

step 2

Find the probability density function of

X

f_X(x)

. Since

X

is uniformly distributed on

[-2,2]

f_X(x) = \frac{1}{4-(-2)} = \frac{1}{6}

for

x \in [-2,2]

step 3

Determine the points where the transformation

Y=(X-1)^2

is not one-to-one. These points occur at

X=1

, where

Y

reaches its minimum

step 4

Solve for

X

in terms of

Y

. We have two solutions:

X = 1 + \sqrt{Y}

and

X = 1 - \sqrt{Y}

step 5

Calculate the derivatives

\frac{dX}{dY}

for each branch. For

X = 1 + \sqrt{Y}

\frac{dX}{dY} = \frac{1}{2\sqrt{Y}}

. For

X = 1 - \sqrt{Y}

\frac{dX}{dY} = -\frac{1}{2\sqrt{Y}}

step 6

Apply the transformation formula for probability density functions. The probability density function of

Y

is given by

f_Y(y) = f_X(x) \left| \frac{dX}{dY} \right|

for each branch of

X

step 7

Combine the results for both branches. Since the transformation is not one-to-one, we must consider both branches of

X

for Y > 0. Thus,

f_Y(y) = f_X(1 + \sqrt{Y}) \left| \frac{1}{2\sqrt{Y}} \right| + f_X(1 - \sqrt{Y}) \left| -\frac{1}{2\sqrt{Y}} \right|

for

y \in (0,9]

step 8

Simplify the probability density function for

Y

. Since

f_X(x)

is constant over its support,

f_Y(y) = \frac{1}{6} \left( \frac{1}{2\sqrt{Y}} + \frac{1}{2\sqrt{Y}} \right) = \frac{1}{6\sqrt{Y}}

for

y \in (0,9]

. For

y=0

f_Y(y) = 0

Answer

f_Y(y) = \begin{cases} \frac{1}{6\sqrt{y}} &amp; \text{for } y \in (0,9] \\ 0 &amp; \text{for } y = 0 \end{cases}

Key Concept

Transformation of Variables in Probability Distributions

Explanation

The probability density function of a transformed variable is found by applying the transformation formula, which involves the original density function and the derivative of the inverse transformation. Since the transformation

Y=(X-1)^2

is not one-to-one, we must consider both branches of

X

when Y > 0. The resulting density function is a piecewise function that accounts for the behavior at

Y=0

and for

Y

in the interval $(0,