CALCU-7. The complete graph K_2 n+1 has no perfect

7. The complete graph

K_{2 n+1}

has no perfect matching since it has an odd number of vertices. Convince yourself that the complete graph

K_{2 n}

has perfect matchings. How many perfect matchings are there in

K_{2 n}

?

Solution by Steps

step 1

Consider the complete graph

K_{2n}

which has

2n

vertices. A perfect matching in a graph is a set of edges without common vertices that covers all vertices of the graph

step 2

The first vertex can be matched with any of the remaining

2n - 1

vertices

step 3

After matching the first vertex, we are left with a complete graph of

2n - 2

vertices

step 4

Now, we can apply this reasoning recursively. The next vertex can be matched with any of the remaining

2n - 3

vertices

step 5

The number of perfect matchings is therefore

(2n - 1) \times (2n - 3) \times \cdots \times 3 \times 1

. This is the product of all odd numbers up to

2n - 1

and is known as the double factorial of

(2n - 1)

step 6

The number of perfect matchings in

K_{2n}

is given by

(2n - 1)!!

Answer

The number of perfect matchings in

K_{2n}

is

(2n - 1)!!

.

Key Concept

Perfect Matchings in Complete Graphs

Explanation

In a complete graph with an even number of vertices

2n

, a perfect matching is a set of

n

edges which covers all vertices exactly once. The number of perfect matchings is given by the double factorial of

(2n - 1)

.

Prove that every tree has at most one perfect matching.

Solution by Steps

step 1

To prove that every tree has at most one perfect matching, we need to understand the structure of a tree and the definition of a perfect matching

step 2

A tree is a connected graph with no cycles, and a perfect matching is a set of edges where every vertex is incident to exactly one edge in the set

step 3

Assume for contradiction that a tree has two different perfect matchings, M1 and M2

step 4

There must be at least one edge in M1 that is not in M2. Let's call this edge e1, and it connects two vertices, say v1 and v2

step 5

Since M2 is also a perfect matching, there must be edges e2 and e3 in M2 that are incident to v1 and v2, respectively, but not in M1

step 6

Now, consider the subgraph formed by the edges e1, e2, and e3. This subgraph contains a cycle, which contradicts the definition of a tree

step 7

Therefore, our assumption is wrong, and a tree cannot have two different perfect matchings

Answer

Every tree has at most one perfect matching.

Key Concept

Perfect Matching in Trees

Explanation

A tree, being cycle-free, cannot accommodate two distinct perfect matchings without creating a cycle, which contradicts its acyclic nature.

1. A random variable

X

has a probability density function given by f_{X}(x)=\frac{2 x+6}{25}, \quad-3<x<2, \quad f_{X}(x)=0 \text { elsewhere. } Let

Y=|X|

. (a) What is

S_{Y}

?

Generated Graph

Solution by Steps

step 1

Identify the support of the random variable

Y=|X|

step 2

Since

X

ranges from

-3

to

2

,

Y

will range from

0

to

3

for

X

negative and from

0

to

2

for

X

positive

step 3

Combine the ranges to find the support

S_Y

for

Y

Answer

S_Y = [0, 3]

Key Concept

Support of a Random Variable

Explanation

The support of a random variable is the set of values where the random variable has a non-zero probability density. For

Y=|X|

, the support is all non-negative values that

X

can take, which are from

0

to

3

.

1. A random variable

X

has a probability density function given by f_{X}(x)=\frac{2 x+6}{25}, \quad-3<x<2, \quad f_{X}(x)=0 \text { elsewhere. } Let

Y=|X|

. (c) Derive the cumulative distribution function of

Y

.

Solution by Steps

step 1

To find the cumulative distribution function (CDF) of

Y

, we first need to understand that

Y = |X|

implies

Y

takes values from

0

to

3

since the support of

X

is from

-3

to

2

step 2

For y < 0, the CDF

F_Y(y) = P(Y \leq y) = 0

since

Y

cannot take negative values

step 3

For 0 \leq y < 2,

F_Y(y) = P(Y \leq y) = P(-y \leq X \leq y)

because

Y = |X|

and

X

can take values between

-y

and

y

step 4

To calculate

P(-y \leq X \leq y)

for 0 \leq y < 2, we integrate the PDF of

X

from

-y

to

y

. This is the sum of the integrals from

-3

to

-y

and from

-y

to

y

since the PDF is symmetric around

0

step 5

Using the result from asksia-ll calculator, the integral from

-3

to

y

is

1/25 (y + 3)^2

. Since the PDF is symmetric and

Y

is non-negative, we need to double the integral from

0

to

y

step 6

Therefore, for 0 \leq y < 2,

F_Y(y) = 2 \times \left[ \frac{1}{25} (y + 3)^2 - \frac{1}{25} (0 + 3)^2 \right]

step 7

For

y \geq 2

,

F_Y(y) = 1

because the entire probability mass has been accumulated

step 8

Simplifying the expression from step 6, we get

F_Y(y) = \frac{2}{25} y^2 + \frac{12}{25} y

for 0 \leq y < 2

Answer

The cumulative distribution function of

Y

is given by:

F_Y(y) = \begin{cases} 0 &amp; \text{for } y &lt; 0, \\ \frac{2}{25} y^2 + \frac{12}{25} y &amp; \text{for } 0 \leq y &lt; 2, \\ 1 &amp; \text{for } y \geq 2. \end{cases}

Key Concept

Cumulative Distribution Function (CDF) of a Random Variable

Explanation

The CDF of a random variable

Y

is found by integrating its probability density function (PDF) over its domain. For

Y = |X|

, we consider the symmetry of

X

's PDF and the non-negativity of

Y

.

1. A random variable

X

has a probability density function given by f_{X}(x)=\frac{2 x+6}{25}, \quad-3<x<2, \quad f_{X}(x)=0 \text { elsewhere. } Let

Y=|X|

. (a) What is

S_{Y}

?

Generated Graph

Solution by Steps

step 1

To find the support

S_Y

of the random variable

Y = |X|

, we consider the range of values that

X

can take

step 2

Since

X

is defined on the interval -3 < x < 2, the absolute value

|X|

will be defined on

0 \leq y \leq 3

step 3

Therefore, the support

S_Y

is the set of all possible values that

Y

can take, which is

[0, 3]

Answer

S_Y = [0, 3]

Key Concept

Support of a Random Variable

Explanation

The support of a random variable is the set of values where the random variable has a non-zero probability density. For

Y = |X|

, the support is the non-negative range of

X

.

1. A random variable

X

has a probability density function given by f_{X}(x)=\frac{2 x+6}{25}, \quad-3<x<2, \quad f_{X}(x)=0 \text { elsewhere. } Let

Y=|X|

. (d) Derive the probability density function of

Y

.

Generated Graph

Solution by Steps

step 1

Identify the range of

X

and the corresponding probability density function (pdf) for

X

step 2

Recognize that

Y = |X|

implies the pdf of

Y

will be symmetrical around

0

step 3

Split the range of

X

into two parts: -3 < x < 0 and 0 \leq x < 2

step 4

For x < 0, use the pdf of

X

to find the pdf of

Y

when Y > 0. This is because

Y = |X| = -X

for x < 0

step 5

For

x \geq 0

, use the pdf of

X

to find the pdf of

Y

when

Y \geq 0

. This is because

Y = |X| = X

for

x \geq 0

step 6

Combine the results from step 4 and step 5 to get the full pdf of

Y

step 7

The pdf of

Y

for y > 0 is the sum of the pdfs of

X

for

x = y

and

x = -y

step 8

Calculate the pdf of

Y

using the formula

f_Y(y) = f_X(y) + f_X(-y)

for 0 < y < 3

step 9

For 0 < y < 2,

f_Y(y) = \frac{2y+6}{25} + \frac{-2y+6}{25}

step 10

Simplify the expression for

f_Y(y)

in the range 0 < y < 2

step 11

For 2 \leq y < 3,

f_Y(y) = \frac{-2y+6}{25}

, since

f_X(y) = 0

for

y \geq 2

step 12

Simplify the expression for

f_Y(y)

in the range 2 \leq y < 3

step 13

Combine the simplified expressions for

f_Y(y)

to write the piecewise function for the pdf of

Y

Answer

The probability density function of

Y

is given by:

f_Y(y) = \begin{cases} \frac{12}{25}, &amp; 0 &lt; y &lt; 2 \\ \frac{-2y+6}{25}, &amp; 2 \leq y &lt; 3 \\ 0, &amp; \text{elsewhere} \end{cases}

Key Concept

Transformation of Random Variables

Explanation

When transforming a random variable, in this case using the absolute value, the pdf of the new variable is derived by considering the contributions from both sides of the original variable's pdf. The new pdf is then a piecewise function that accounts for the different behaviors in different intervals of the original variable's domain.

1. A random variable

X

has a probability density function given by f_{X}(x)=\frac{2 x+6}{25}, \quad-3<x<2, \quad f_{X}(x)=0 \text { elsewhere. } Let

Y=|X|

. ) Comment on the continuity of the cdf and the pdf of

Y

.

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