Consider discrete random variable X and Y with probabilities as follows:
$P(X=0\text{ and }Y=0)=\frac{1}{4}$
$P(X=1\text{ and }Y=1)=\frac{1}{8}$
$P(X=0\text{ and }Y=1)=\frac{1}{2}$
$P(X=1\text{ and }Y=1)=\frac{1}{8}$
Given $X=1$, the expected value of Y is

The question asks for the expected value of the discrete random variable Y, given that the random variable X is equal to 1. This is denoted as $E[Y|X=1]$.

Understanding Joint Probabilities

We are given the following joint probabilities for discrete random variables X and Y. Note: The fourth listed probability, $P(X=1\text{ and }Y=1)=\frac{1}{8}$, appears to be a duplicate or typo. To form a valid probability distribution where the sum of probabilities for all possible outcomes is 1, we assume the distinct probabilities cover the four possible pairs of outcomes for binary variables X and Y (0 or 1). Based on this, we infer the complete distribution:

- $P(X=0,Y=0)=\frac{1}{4}$
- $P(X=0,Y=1)=\frac{1}{2}$
- $P(X=1,Y=0)=\frac{1}{8}$ (Inferred to complete the distribution)
- $P(X=1,Y=1)=\frac{1}{8}$

Let's verify the sum: $\frac{1}{4}+\frac{1}{2}+\frac{1}{8}+\frac{1}{8}=\frac{2}{8}+\frac{4}{8}+\frac{1}{8}+\frac{1}{8}=\frac{8}{8}=1$. The distribution is valid.

Calculating Marginal Probability P(X=1)

To find the conditional expectation $E[Y|X=1]$, we first need the marginal probability of the condition, $P(X=1)$. This is calculated by summing the joint probabilities where $X=1$:

$P(X=1)=P(X=1,Y=0)+P(X=1,Y=1)$

Substituting the known values:

$P(X=1)=\frac{1}{8}+\frac{1}{8}=\frac{2}{8}=\frac{1}{4}$

Determining Conditional Probabilities P(Y=y | X=1)

Next, we determine the conditional probabilities of Y given X=1. The formula for conditional probability is $P(A|B)=\frac{P(A\cap B)}{P(B)}$. Thus:

$P(Y=y|X=1)=\frac{P(X=1,Y=y)}{P(X=1)}$

For $Y=0$:

$P(Y=0|X=1)=\frac{P(X=1,Y=0)}{P(X=1)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{8}\times 4=\frac{4}{8}=\frac{1}{2}$

For $Y=1$:

$P(Y=1|X=1)=\frac{P(X=1,Y=1)}{P(X=1)}=\frac{\frac{1}{8}}{\frac{1}{4}}=\frac{1}{8}\times 4=\frac{4}{8}=\frac{1}{2}$

Computing the Conditional Expected Value

The expected value of Y given X=1 is calculated using the conditional probabilities:

$E[Y|X=1]={\sum }_{y\in \{0,1\}}y\cdot P(Y=y|X=1)$

$E[Y|X=1]=(0\cdot P(Y=0|X=1))+(1\cdot P(Y=1|X=1))$

Substituting the calculated conditional probabilities:

$E[Y|X=1]=(0\cdot \frac{1}{2})+(1\cdot \frac{1}{2})$

$E[Y|X=1]=0+\frac{1}{2}=\frac{1}{2}$