Expected value, dispersion and correlation coefficient of losses during the day

Question

Every day a kid puts $7$ £1 coins and $9$ £2 coins in his pocket. Each day, in the morning he loses $3$ coins and in the evening of the same day he loses $5$ coins. Find his morning and evening losses' expected value $E(X)$, dispersion $D(X)$ and the correlation coefficient.

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So, calculating his morning losses is fairly simple:

If he loses 3 coins, his losses must be $≥3$ and $≤6$ since he can lose both £1 coins and £2 coins.

If his loss is £3 in total, he must have lost $3$ £1 coins: $C(7, 3) = 35$,

If his loss is £4 in total, he must have lost $2$ £1 coins and $1$ £2 coin: $C(7, 2) · C(9, 1) = 189$,

If his loss is £5 in total, he must have lost $1$ £1 coin and $2$ £2 coins: $C(7, 1) · C(9, 2) = 252$,

If his loss is £6 in total, he must have lost $3$ £2 coins: $C(9, 3) = 84$.

There are $C(16, 3) = 560$ different combinations of him losing 3 coins.

So, $E(X) = 3 · \frac{35}{560} + 4 · \frac{189}{560} + 5 · \frac{252}{560} + 6 · \frac{84}{560} = 4.6875$

And then accordingly, $D(X) = 3^2 · \frac{35}{560} + 4^2 · \frac{189}{560} + 5^2 · \frac{252}{560} + 6^2 · \frac{84}{560} - (4.6875)^2 ≈ 0.6398 $

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However, I can't think of a way how to calculate the evening losses for multiple reasons.

For example, if he has $7$ £1 coins in total and loses $3$ £1 coins in the morning, he can't lose 5 of them in the evening.

How do I account for this, and, most importantly - how do I calculate the correlation coefficient?

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Edit based on the replies:

When it comes to calculating the evening losses, the morning losses must be accounted for, too.

For example, that's my calculation for the loss of £5:

• Given that he lost $3$ £1 coins in the morning, he has $4$x £1 and $9$x £2 coins.

If his loss is £5 in total, he could have lost:

$3$ £1 coins and $1$ £2 coin or $1$ £1 coin and $2$ £2 coins, $C(4, 3) ⋅ C(9, 1) + C(4,1) ⋅ C(9, 2) = 180$

There are $C(13, 5)$ ways to lose $5$ coins out of leftover $13$ and the probability of him losing $3$ £1 coins in the morning is $\frac{35}{560}$

Thus, $\frac{35}{560} ⋅ \frac{180}{1287} = \frac{5}{572}$

• Given that he lost $2$ £1 coins and $1$ £2 coin in the morning, he has $5$x £1 and $8$x £2 coins.

If his loss is £5 in total, he could have lost:

$5$ £1 coins or $3$ £1 coins and $1$ £2 coin or $1$ £1 coin and $2$ £2 coins, $C(5, 5) + C(5, 3) ⋅ C(8, 1) + C(5,1) ⋅ C(8, 2) = 221$

Thus, $\frac{35}{560} ⋅ \frac{221}{1287} = \frac{51}{880}$

• Given that he lost $1$ £1 coin and $2$ £2 coins in the morning, he has $6$x £1 and $7$x £2 coins.

If his loss is £5 in total, he could have lost:

$5$ £1 coins or $3$ £1 coins and $1$ £2 coin or $1$ £1 coin and $2$ £2 coins, $C(6, 5) + C(6, 3) ⋅ C(7, 1) + C(6,1) ⋅ C(7, 2) = 272$

Thus, $\frac{35}{560} ⋅ \frac{272}{1287} = \frac{68}{715}$

• Given that he lost $3$ £2 coins in the morning, he has $7$x £1 and $6$x £2 coins.

If his loss is £5 in total, he could have lost:

$5$ £1 coins or $3$ £1 coins and $1$ £2 coin or $1$ £1 coin and $2$ £2 coins, $C(7, 5) + C(7, 3) ⋅ C(6, 1) + C(7,1) ⋅ C(6, 2) = 336$

Thus, $\frac{35}{560} ⋅ \frac{336}{1287} = \frac{28}{715}$

So, the probability of his loss of £5 in the evening is $\frac{5}{572} + \frac{51}{880} + \frac{68}{715} + \frac{28}{715} = \frac{209}{1040}$

and I would go on with 6, 7, 8, 9, 10 in quite a similar manner.

Answer 1

Try using conditional expectation:

$Y$...pounds lost in the evening

Calculate $E(Y|X=k), k=3,4,5,6$

Then: $E(Y)=\sum_{k=3}^6E(Y|X=k)P(X=k)$

For the correlation coefficient, use the formula:

$r(X,Y)=\frac{K(X,Y)}{\sqrt{D(X)D(Y)}}$, where $K(X,Y)=E(XY)-E(X)E(Y)$

Answer 2

I am not sure about the answer. But as the above answer suggest using conditional expectation we will get the answer. But the calculation would be tedious. I have this idea.

If this equation holds true; $$ \text{E}[ \text{losing 8 coins} ] = \text{E}[ \text{losing 3 coins}] + \text{E} [\text{losing 5 coins}]$$

So this will make a calculation litter easier than the method using conditional probabilities. I don't know whether this above equation holds true.[I have no access to comment so I wrote an answer just correct me if I am wrong].

This works for $$ \text{E}[2] = 2 * \text{E}[1] $$ $$\frac{\binom{9}2}{\binom{16}2}*4+\frac{\binom{9}1 *\binom{7}1 }{\binom{16}2}*3 + \frac{\binom{7}2}{\binom{16}2}*2=\frac{375}{120}=2*(\frac{7}{16}*1+\frac{9}{16}*2 $$)