X is uniformly distributed on the interval [1,4] and Y = sqrt(X)

Question

Suppose that X is uniformly distributed on the interval [1,4], and that Y = sqrt(X). Evaluate E(Y) and Var(Y).

I know the formulas to get the expected value and variance of a uniform distribution. I'm just not sure how to get from X to Y. Can I get the expected value and variance of Y by taking the square root of E(X) and Var(X)? I'm not sure how else to approach this.

Write out the formula and then do change of variables to get the integral in terms of X's sample space. — Paul, Jul 16 '15 at 20:16
To get started: $E(Y) = E(\sqrt{X}) = \frac{1}{3}\int_1^4 \sqrt{x} dx$ — , Jul 16 '15 at 20:17

score 2 · Answer 1 · answered Jul 16 '15 at 20:23

Suppose you know the density of a random variable $X$. You have $Y=\text{some function of }X$, for example $Y=X^3 + \cos(\sqrt X)$.

You could find the density of the random variable $Y$ and then write $$ \operatorname{E}(Y) = \int_\text{?}^\text{?} y f_Y(y)\,dy. $$ But you can also do it the following way, which is often less onerous: $$ \operatorname{E}(Y) = \int_\text{?}^\text{?} (x^3+\cos(\sqrt x)) f_X(x)\,dx. $$ Sometimes this is called "the law of the unconscious statistician". Google that term.

The bounds of integration for the first integral would in general be different from those of the second. In your example, as $x$ goes from $1$ to $4$ then $\sqrt x$ goes from $1$ to $2$. But you don't need that if you use the second integral above.

X is uniformly distributed on the interval [1,4] and Y = sqrt(X)

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