0
$\begingroup$

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 $4X-3$

$\endgroup$
1
$\begingroup$

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 $Y$:

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ thank you. The first expression will evaluate to my solution $\alpha = 4$ $\beta = -3$ then we have $4\cdot \frac{1}{3} -3 = -\frac{5}{3}$ For the variance I get $16\cdot \frac{1}{9} = \frac{16}{9}$ due to the fact $Var[X]=\frac{1}{3^2}$. $\endgroup$ – Anil Dec 5 '17 at 14:03
  • $\begingroup$ But what about cdf and pdf? $\endgroup$ – Anil Dec 5 '17 at 14:03
  • $\begingroup$ @Anil I’m really sorry, completely overlooked that part. I will change my reply in a few $\endgroup$ – caverac Dec 5 '17 at 14:06
  • $\begingroup$ @Anil I just updated my answer $\endgroup$ – caverac Dec 5 '17 at 17:18
  • $\begingroup$ thank you for your edit, but I don't know what to use for $f_X(x)$ and $f_Y(y)$ Is it something like $f_X(x)=3e^{-3x}$ ? $\endgroup$ – Anil Dec 5 '17 at 23:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.