Question

Let \(X\) be a continuous random variable with cdf \(F(x)\). Suppose \(Y=X+\Delta\), where \(\Delta>0\). Show that \(Y\) is stochastically larger than \(X\).

Short answer

Yes, \(Y = X + \Delta\), where \(\Delta > 0\), is stochastically larger than \(X\) as its cumulative distribution function \(F_Y(y)\) lies everywhere to the right of the cumulative distribution function \(F_X(x)\) for all \(x\) and \(y\) in their domain.

Step-by-step solution

Remember the definition of a CDF

The cumulative distribution function (CDF) of a random variable \(X\) is defined as \(F(x) = P(X \leq x)\). It gives the probability that the random variable \(X\) takes a value less than or equal to \(x\).

Translate Y in terms of X

Given \(Y = X + \Delta\), we can express this in terms of \(X\) as \(X = Y - \Delta\). This will help us connect the CDFs of \(Y\) and \(X\).

Express CDF of Y

Express the CDF of \(Y\) in terms of the CDF of \(X\). We have \(F_Y(y) = P(Y \leq y) = P(X + \Delta \leq y) = P(X \leq y - \Delta) = F_X(y-\Delta)\). Notice here \(F_Y(y)\) is equivalent to shifting \(F_X(x)\) to the right by \(\Delta\) units.

Compare the CDFs of X and Y

Because \(\Delta > 0\), it is clear that \(Y = X + \Delta\) implies that \(F_Y(y) = F_X(y-\Delta) \leq F_X(x)\) for all \(x\) and \(y\). Thus, \(Y\) is stochastically larger than \(X\).