My Math Forum Multiplying a Random Variable, poisson dists and variance

 November 9th, 2011, 04:56 PM #1 Senior Member   Joined: Sep 2009 Posts: 251 Thanks: 0 Multiplying a Random Variable, poisson dists and variance OK, I either have one simple question "How do you multiply a random variable by a number?", or utter confusion. I think it's all centered around the weird notations in Probability. A random variable, X, is a coin flip variable (that is, it's either a 0 or a 1), and the Poisson distribution's Probability Mass D(something) (PMD) is $p_X(x)=\frac{e^{-\lambda}\lambda^{x}}{x!}$. I have that Expected Value and Variance are $E[X]=\sum_{\forall x}{x \cdot p_{X}(x)} \\ var(X)=E[(X-E[x])^2].$ I'm also told that the expected value and variance of a Poisson Distribution is lambda (which makes sense). I have answers to my friend's quiz which says that, if lambda=.2 of Poisson variable X, then $E[X]=.2 \\ var(X)=.2 \\ E[3X+.2]=1.1 \\ var(3X)=1.8.$ She thinks that E[3X+2]=3(.2)+.2=1.1, which I know is wrong, but can't convince her otherwise. Neither of us understand why the var(3X)=1.8. But my question is not just how did they those answers. I really want to know how do you multiply 3 times a random variable, X (and then how do you add 2 to that)? The variable is not a number, it's an idea - it's the number of occurrences of an event (e.g. coin flips heads). After I get that, I need to apply it to finding the EV and variance of 3X, 3X+2, etc. I'm also trying to figure out how they calculate that E[poisson variable]=lambda and variance(poisson variable)=lambda. Thanks, everyone.
 November 9th, 2011, 08:39 PM #2 Joined: Oct 2011 Posts: 26 Thanks: 0 Re: Multiplying a Random Variable, poisson dists and varianc Rules: -Adding a constant to a random variable increases the expectation of the random variable by that constant $E(X+C)= E(X)+C$ -Multiplying a random variable by a constant means that the expected value of that random variable is multiplied by the constant $E(CX)= CE(X)$ -Adding a constant to a random variable does not change the variance of the random variable $Var(X+C)= Var(X)$ -Multiplying a random variable by a constant multiplies the variance of the random variable by the square of that constant $Var(CX)= C^2Var(X)$ Memorise these rules and you'll be able to answer exam questions. Do some googling if you want to see proof of these rules and really understand them. In your example, you have $E(3X+.2)= E(3X)+.2 = 3E(X)+.2 = 3(.2)+.2 = .8$ $Var(3X)= 3^2Var(x) = 9(.2) = 1.8$ It looks like our answer for the expectation doesn'td agree. Now I'm confused... Anyway, something else: you say that the it "makes sense" that the expectation and variance of a Poisson distribution are lambda, but as far as I know that's at least as complicated as the rules for the expectation and variance. You might want to look here for an explanation of why the expectation of a Poisson distribution is lambda:
November 9th, 2011, 09:11 PM   #3
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Re: Multiplying a Random Variable, poisson dists and varianc

Quote:
 Originally Posted by newguy123 Rules: -Adding a constant to a random variable increases the expectation of the random variable by that constant $E(X+C)= E(X)+C$ -Multiplying a random variable by a constant means that the expected value of that random variable is multiplied by the constant $E(CX)= CE(X)$ -Adding a constant to a random variable does not change the variance of the random variable $Var(X+C)= Var(X)$ -Multiplying a random variable by a constant multiplies the variance of the random variable by the square of that constant $Var(CX)= C^2Var(X)$ Memorise these rules and you'll be able to answer exam questions. Do some googling if you want to see proof of these rules and really understand them. In your example, you have
Ahhh. That clears everything up. Good answer.
The first three will be easy to memorize. I will have to look up a proof of that variance one. (Edit: I retract that last sentence - it was easy to figure out.)

Quote:
 Originally Posted by newguy123 $E(3X+.2)= E(3X)+.2 = 3E(X)+.2 = 3(.2)+.2 = .8$ $Var(3X)= 3^2Var(x) = 9(.2) = 1.8$ It looks like our answer for the expectation doesn'td agree. Now I'm confused...
The confusion is that my OP is almost certainly copied wrong. My friend was late for her class and I had to drive her back to school. I copied the problems down in the hurry. Lambda, the probability that I made a write-o (that's a type-o made with pencil and paper) is somewhere north of .7 for each word I rushed to write (you can use a Poisson distribution to determine the likelihood that my original post was completely accurate ). So don't be confused.

Quote:
 Originally Posted by newguy123 Anyway, something else: you say that the it "makes sense" that the expectation and variance of a Poisson distribution are lambda, but as far as I know that's at least as complicated as the rules for the expectation and variance. You might want to look here for an explanation of why the expectation of a Poisson distribution is lambda:
I mean it "makes sense" in that it's an intuitive answer. If lambda is the probability of something happening, and they're all equal weighted, then it just "seems" like the right answer for expected. I don't mean that I can calculate it or genuinely understand it.

In fact, I first started to try and calculate the expected of a poisson using the integral
$E[X]=\int{x\ p_{X}(x)}\ dx=\int{x\ \frac{e^{-\lambda}\ \lambda^{x}}{x!}}\ dx$
but I could not figure out how to do it and I couldn't get Maple to do it (I can never get Maple to do anything useful). Instead, I looked in another textbook and found it.

Thanks again, newbie.

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# scalar multiplication of poisson variable

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