Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I want to define a new random variable $f$ as a function of a normal random variable $v$: $$f(v)=\begin{cases}C&\text{if } v\ge C\\ \gamma v &\text{otherwise}\end{cases}$$

where $v\sim N(v_0,\sigma_v)$ and $C$ is constant.

How should I compute the variance of $f$? Can anyone help me? Thanks, PM

share|improve this question
1  
I don't think this is a valid density function. How can you have finite probability on the interval from $C$ to infinity? –  trb456 Feb 10 '13 at 12:21
1  
@trb456: As $v$ is normally distributed, the variance of $f$ will be finite. –  Ron Gordon Feb 10 '13 at 12:53
    
@user61677: As my comments are suggesting, I think some editing to your question would help clarify what you want. Instead of your function f, which I am likely mistaking for a pdf, do instead mean you want to define a new random variable as a function of an existing normal random variable? This may seem picky, but I think it would be more clear. –  trb456 Feb 10 '13 at 13:29
    
Twice the post mentions $f$ and twice it should read $f(v)$. (Note furthermore that random variables are most often denoted by capital letters, in which case the random variable of interest would be $f(V)$.) –  Did Feb 10 '13 at 14:10

2 Answers 2

I think the way you want to go about this is to express the expected value of $f$ as

$$E[f(V)] = \frac{\gamma}{\sqrt{2 \pi} \sigma_0} \int_{-\infty}^C dv \: v \exp{\left [ - \frac{(v-v_0)^2}{2 \sigma_0^2} \right ]} + \frac{C}{\sqrt{2 \pi} \sigma_0} \int_{C}^{\infty} dv \: \exp{\left [ - \frac{(v-v_0)^2}{2 \sigma_0^2} \right ]} $$

where $V$ is the normally distributed random variable. Also,

$$E[f(V)^2] = \frac{\gamma^2}{\sqrt{2 \pi} \sigma_0} \int_{-\infty}^C dv \: v^2 \exp{\left [ - \frac{(v-v_0)^2}{2 \sigma_0^2} \right ]} + \frac{C^2}{\sqrt{2 \pi} \sigma_0} \int_{C}^{\infty} dv \: \exp{\left [ - \frac{(v-v_0)^2}{2 \sigma_0^2} \right ]} $$

The variance is

$$\mathrm{Var}[f(V)] = E[f(V)^2] - E[f(V)]^2$$

share|improve this answer
    
I'm sorry, but what you are integrating is not the OP's function, which I again reiterate is not a valid density function as written. –  trb456 Feb 10 '13 at 13:03
    
@trb456: Where did the OP say that $f$ is a density function? $f$ is not a density function, but rather a function of the random variable $V$. –  Ron Gordon Feb 10 '13 at 13:05
    
@rlgordonna: The new random variable still needs a density function, and what you are integrating is not it. At a minimum, you have the support backwards (constant above $C$ as written). But more seriously, what is the pdf when the normal random variable exceeds $C$? You seem to want to assume it's normally distributed above $C$, and you are likely be right that this is what the OP meant, but if so then this could be better expressed than currently. –  trb456 Feb 10 '13 at 13:26
    
@trb456: thanks for catching my error. As for the density function of $f(V)$, I am not assuming anything that the OP hasn't stated. –  Ron Gordon Feb 10 '13 at 13:30
1  
Thanks a lot!! f is not a density function, but a function of the random variable v. –  user61677 Feb 10 '13 at 13:30

You can use the results for a truncated normal distribution. I assume $\sigma_v$ is a standard deviation rather than a variance.

Let $\beta = \dfrac{C-v_0}{\sigma_v}$, $b=\phi(\beta)$ and $B=\Phi(\beta)$ using the PDF and CDF of a standard normal distribution. Then the first moment about zero could be something like $$E[f] = B \gamma \left( v_0 - \frac{b}{B} \sigma_v \right) +(1-B)C$$

which you can simplify.

You can then work out the second moment in a similar way and thus the variance of $f$, but I will leave that for you. For what it is worth, my attempt at the second moment (before I decided to stop, so unchecked) was

$$ B \gamma^2 \left(\sigma_v \left( 1 - \frac{b \beta}{ B} -\frac{b^2}{B^2} \right) + \left(v_0 - \frac{b}{B} \sigma_v\right)^2\right) +(1-B)C^2$$ and you just need to subtract the square of the first moment to give the variance.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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