Find $E(Y)$ such that $F_Y(y)= (1 - 0. 3e^{−0.5} + 0. 6e^{−0.25y})$

Question

Find $E(Y)$ such that the cumulative density function of $Y$ is: $$F_Y(y)= (1 - 0. 3e^{−0.5} + 0. 6e^{−0.25y})$$

Can someone help me figure out how to calculate the mean?

Answer 1

What you have provided is likely 1 minus a cumulative distribution function but with at least one typo.

You gave

$$0.3 e^{-0.5 y}+0.6 e^{-0.25 y}$$

A plot of that function is

This function needs to be 1 when $y=0$. To obtain that you probably need

$$0.3 e^{-0.5 y}+0.7 e^{-0.25 y}$$

or

$$0.4 e^{-0.5 y}+0.6 e^{-0.25 y}$$

or in general the weights need to be between 0 and 1 and sum to 1:

$$w e^{-0.5 y}+(1-w) e^{-0.25 y}$$

Suppose we take

$$Pr(Y>y)=0.4 e^{-0.5 y}+0.6 e^{-0.25 y}$$

Then the cumulative distribution function is

$$Pr(Y\leq y)=1-0.4 e^{-0.5 y}-0.6 e^{-0.25 y}$$

The probability density function is

$$f(y)=0.2 e^{-0.5 y}+0.15 e^{-0.25 y}$$

The mean is given by

$$\mu =\int_0^{\infty } y \left(0.2 e^{-0.5 y}+0.15 e^{-0.25 y}\right) \, dy=3.2 $$

and the variance

$$\sigma^2=\int_0^{\infty }(y-\mu )^2 \left(0.2 e^{-0.5 y}+0.15 e^{-0.25 y}\right) \, dy=12.16$$

Answer 2

Note: the probability density function that you have provided does not appear to describe a valid probability distribution. Instead, I will explain the general process of how to solve your problem for any suitable continuous distribution of a similar form.

The probability density function for the random variable Y that you provided is:

$$f_Y(y)= P(Y=y)=0. 3e^{−0.5} + 0. 6e^{−0.25y}$$

We will assume that this is a valid distribution for the purposes of illustrating the method to solve your problem. Note that since Y is continuous, we define its expectation by the integral:

$$E[Y]=\int_{y} yP(Y=y)dy = \int_{y} 0.3ye^{−0.5} + 0. 6ye^{−0.25y}dy=\mu$$

and similarly, the variance is defined:

$$Var(Y) = E(Y^2)-E(Y)^2=\int_{y} y^2P(Y=y)dy=\Big{(}\int_{y} 0.3y^2e^{−0.5} + 0. 6y^2e^{−0.25y}dy\Big{)}-\mu^2$$

If it turns out that you provided us with wrong probability mass function in the question, then this process can be easily adjusted to suit the new distribution.