problem 59

(a)

WLOG, let

m_{1} > m

be the second median of

X

. Then, by the definition of medians,

P (X \leq m) \geq \frac{1}{2}

and

P (X \geq m_{1}) \geq \frac{1}{2}

. Then,

P (X \in (m, m_{1})) = 0

. If

m_{1} > m + 1

, then there exists an

m_{2} \in (m, m_{1})

, such that

P (X = m_{2}) = 0

. This implies that

m_{2} = 1

, since that is the only value of

X

with probability

0

. However, then

m < 1

, which precludes

m

from being a median. Thus,

m_{1}

must be

1 + m

. Since we know

23

to be a median of

X

, we need to check whether

22

24

are medians of

X

. Computation via the CDF of

X

shows that niether

22

, nor

24

are medians. Hence,

23

is the only median of

X

(b)

Let

I_{j}

be the indicator variable for the event

X \geq j

. Notice that the event

X = k

(the first occurance of a birthday match happens when there are

k

people) implies that

I_{j} = 1

for

j \leq k

and vice versa. Thus,

X = 366 \sum j = 1 I_{j}

Then,

E (X) = 366 \sum j = 1 P (I_{j} = 1) = 1 + 1 + 366 \sum j = 3 P (I_{j} = 1) = 2 + 366 \sum j = 3 p_{j}

(c)

2 + 22.61659 = 24.61659

(d)

E (X^{2}) = E (I_{1}^{2} + \dots + I_{366}^{2} + 2 \sum_{j = 2}^{366} \sum_{i = 1}^{j - 1} (I_{i} I_{j}))

. Note that

I_{i}^{2} = I_{i}

and

I_{i} I_{j} = I_{j}

for

i < j

. Thus,

\begin{matrix} E (X^{2}) & = E (I_{1} + \dots + I_{366} + 2 \sum_{j = 2}^{366} \sum_{i = 1}^{j - 1} I_{j}) = 2 + \sum_{j = 3}^{366} p_{j} + 2 \sum_{j = 2}^{366} ((j - 1) E (I_{j})) = 2 + \sum_{j = 3}^{366} p_{j} + 2 \sum_{j = 2}^{366} ((j - 1) p_{j}) \approx 754.61659 \end{matrix}

$Var (X) \approx 754.61659 - {(E (X))}^{2} \approx 754.61659 - 605.98 = 148.63659$ .