Remembering the mean and variance of Poisson vs Exponential I am having the P-Exam for actuaries on September and I am trying to work on my details.
One of the troubles that I have is to remember the difference between Poisson and Exponential distribution.
I understand how to find the mean and variance for both distribution, but it would help me save a lot of time if I did not have to derive them every single time.  Therefore, I tried to memorize their important qualities such as mean, variance, pdf(pmf) etc.
However, assuming that $\lambda$ is the mean for both Poisson and Exponential distributions, I always forget which variance was $\lambda$ or $\lambda^2$.
Does anyone know a simple trick, or an easy way to remember which?
 A: A Poisson distribution models a (rare) event (e.g. plane crash, winning the lottery, etc.), with a known average rate of occurence, $\lambda$ (e.g. 4 people won the lottery last year, so $\lambda=4$).
The weird thing about the Poisson distribution is that the mean is the same as the variance!
Essentially, a Poisson distribution tells you how likely it is that a rare event occurs, if we know the average rate.
If $X \sim \rm{Po}(\lambda),$ then $\color{red}{\boxed{{E[X]=\rm{Var}[X]=\lambda}}}$.
$$\Pr[X=r]=\frac{e^{-\lambda} \times\lambda^r}{r!}.$$
Please, for God's sake, don't try to derive these! It's pretty long and a waste of time. Just learn them!
Finally, the Poisson distribution is discrete.
...
Have you ever heard of the geometric distribution? This is about how many (Bernoulli- two outcomes) trials we have to perform in order to get a certain result. e.g. we can calculate the probability that it takes 10 tosses of a coin to get a heads.
The exponential distribution is like a continuous version of the geometric distribution, so, instead of tossing a coin loads of times until heads comes up, think of it as how long it takes (time is a continuous variable) until an event happens, e.g. how long it takes until a phone rings in a call centre.
The interesting property about the exponential distribution is that it's memoryless.
Say you've been waiting for a bus for 15 minutes. The probability that the bus arrives in 10 more minutes (25 minutes in total) is the same as if you'd just got there and the bus  comes 10 minutes later.
Now, important formulae:
If $X \sim \rm{Exp} (\lambda)$:
$\color{green}{\boxed{E[X]=\lambda^{-1}, \rm Var[X]=\lambda^{-2}}}$
pdf: $\Pr[X \approx x]=f(x)=\lambda e^{-\lambda x}$
The probability that you wait between $\alpha$ and $\beta$ minutes for your bus is:
$$\int\limits_{\alpha}^{\beta}\lambda e^{-\lambda x}dx.$$
The exponential distribution is a continuous distribution. 
A: For the exponential distribution, the mean $\theta$ is a scale parameter.  That means changing the value of $\theta$ is just rescaling --- essentially changing the units of measurement from inches to meters or whatever. In other words, if $X$ is exponentially distributed with mean $1$, then $Y=\theta X$ is exponentially distributed with mean $\theta$.  If you multiply an exponentially distributed random variable by $3$, for example, you get another exponentially distributed random variable.
Recall that $\operatorname{var}(Y) = \operatorname{var}(3X)=3^2\operatorname{var}(X)$ or in general $\operatorname{var}(Y) = \operatorname{var}(\theta X)=\theta^2\operatorname{var}(X)$.  So the variance must be proportional to the square of the scale parameter.
For the Poisson distribution, the expected value $\lambda$ is not a scale parameter.  If it were, then making $\lambda$ three times as big would change the support of the distribution from $\{0,1,2,3,4,5,\ldots\}$ to $\{0,3,6,9,12,15,\ldots\}$, making all the numbers $3$ times as big.  That is rescaling.  Obviously that doesn't happen: we get another Poisson distribution with an expected value $3$ times as big but with the same set as the support.  So multiplying by $\lambda$ is not rescaling, so we don't have a reason to think the variance is proportional to the square of $\lambda$.
A: When I forget what is going on, I move to the concept of rate.  
Both distributions for a rate of $1$ have a mean of $1$, a variance of $1$, and a standard deviation of $1$. Now double the rate. 


*

*The exponential distribution with double the rate sees everything happen potentially twice as quickly, so the mean halves and since this is equivalent to simply scaling time the standard deviation also halves (making the variance the square of this i.e. a quarter of what it was before).

*The Poisson distribution with double the rate in effect sees twice as many things potentially happen, in effect the sum of two independent Poisson distributions of rate $1$, so the mean becomes $1+1=2$ and the variance becomes $1+1=2$ (making the standard deviation become $\sqrt 2$).
This thought process reminds me that the mean of an exponential distribution is equal to the standard deviation (and the reciprocal of the rate), while the mean of a Poisson distribution is equal to the variance (and the rate).
A: Before I answer your question, I feel I should make a very important point with respect to the actuarial exams.  If you for whatever reason find it difficult to remember which distribution has which variance, I must warn you that the amount of memorization required for subsequent exams will increase, and it will increase dramatically.  Exam 3L/MLC and Exam 4/C, for instance, make Exam 1/P look like basic arithmetic.  Past the preliminary examinations, memorization becomes even more important as you will need to remember various regulatory and accounting issues encountered in actuarial practice, and unlike the preliminary exams for which one could at least derive forgotten results from other known facts, you just have to know these regulations.
That said, the simplest way to remember the differences is to distinguish the exponential distribution from the Poisson by using the scale parametrization for the former, rather than the rate parametrization.  You should learn it this way anyway; Exam 4/C will frequently use a scale parametrization for the exponential and gamma distributions.  Then you will not get confused:  If $X \sim {\rm Exponential}(\theta)$, then $$f_X(x) = \frac{1}{\theta}e^{-x/\theta}, \quad x > 0,$$ and $${\rm E}[X] = \theta, \quad {\rm Var}[X] = \theta^2.$$  The Poisson is always parametrized in terms of the rate $\lambda$:  $$\Pr[N = k] = e^{-\lambda}\frac{\lambda^k}{k!}, \quad k = 0, 1, 2, \ldots, $$ and $${\rm E}[N] = {\rm Var}[N] = \lambda.$$  But as I pointed out earlier, if you have difficulty getting this straight, then you are in for an extremely difficult and painful time with other exams.  That is merely fact and observation, not condescension nor an attempt at intimidation.
