Expectation, variance etc. for exponential distribution

Question

Suppose $X\sim exp(3)$

Now I have to consider $4X-3$ and determine i) expectation ii) variance iii) cdf iv) pdf

We know that $$\mathbb{E}[Y]=\frac{1}{\lambda}$$

For my case I have to consider $\mathbb{E}[X]=\frac{1}{3}$. Does that mean I can plug $\frac{1}{3}$ into the equation like that or will be this wrong? $\mathbb{E}[4X-3]\stackrel{?}{=} 4\cdot\frac{1}{3}-3=-\frac{5}{3}$

The same thing for the variance

$$Var[Y]=\frac{1}{\lambda^2}$$

For my case $Var[X]=\frac{1}{3^2}=\frac{1}{9}$ Can I plug $\frac{1}{9}$ into $Var[4X-3]\stackrel{?}{=}4\cdot\frac{1}{9}-3=-\frac{23}{9}$

Or will be this approach completely wrong? These were my initial thoughts.

For (1) probability density function and (2) cumulative distribution function these following equations are given, but I don't know how to use them for my case

$$(1) \ \ f(b)db=\lambda e^{\lambda b}db$$ $$ (2) \ \ F(b)=1-\mathbb{P}(Y\ge b)=1-e^{-\lambda b}$$

$$\lambda=3$$ $$(1) \ \ f(b)db=3 e^{3 b}db$$ $$ (2) \ \ F(b)=1-\mathbb{P}(Y\ge b)=1-e^{-3 b}$$

Will this work out because I don't know how to combine this with $4X-3$

Answer 1

What you know is that $\lambda = 3$, so that

$$ \mathbb{E}[X] = 1/\lambda = 1 / 3 $$

You can then use the expressions

$$ \mathbb{E}[\alpha X + \beta] = \alpha \mathbb{E}[X] + \beta $$

and

$$ \mathbb{V}{\rm ar}[\alpha X + \beta] = \alpha^2\mathbb{V}{\rm ar}[X] $$

As for the pdf you can use the fact that

$$ f_X(x) dx = f_Y(y) dy $$

Where $f_X$ is the pdf of $X$, same for $Y$. Solve for the pdf of $Y$:

$$ f_Y(y) = f_X(x)\frac{dx}{dy} $$