Discrete Probability: Expected value of random variable

Question

Question: You flip a fair red coin once, and you flip a fair blue coin once, independently of each other.

Define the random variables:

X = {1 if the red coin flip resulted in heads, 0 if the red coin flip resulted in tails}

Y = {1 if the blue coin flip resulted in heads, 0 if the blue coin flip resulted in tails}

and Z = min(X,Y)

What is the expected value E(Z) of the random variable Z?

Answer: 0.25

Attempt:

For random variable x=1 correlates to pr(x) = 0.5 and x = 0 correlates to pr(x) = 0.5. Same with the y's.

Z=min(X,Y) = would just be either since their the same?

So, using the expected value formula I get that (0.5*1) + (0.5*0) = 0.5

Answer 1

$Z$ is an indicator variable and it takes value $1$ when both get heads.

Hence $$E[Z]=P(Z=1) = 0.5^2=0.25$$

Remark:

The assumption that $Z$ would follows the same distribution as $X$ or $Y$ is false.

Answer 2

It is a fallacy to assume that $X$ and $Y$ are the “same” because their probability distributions have the same weights.

If the coins are independent, then by the definition of independence, $$ P(X=1 \text{ and } Y=1) = P(X=1) P(Y=1). $$ There are three other possible outcomes, including

• $X=1$ and $Y=0$;
• $X=0$ and $Y=1$;
• $X=0$ and $Y=0.$

The probability of each of those events can be worked out similarly to $P(X=1 \text{ and } Y=1).$

In all of those cases the final probability comes out to $0.25.$ But the case “$X=1$ and $Y=1$” is the only case in which $\min(X,Y)=1.$ In every other case the result is $0.$ So altogether you have a $0.75$ chance of $0,$ not $0.5.$