1
$\begingroup$

Suppose that $X_1$, $X_2$, ..., $X_n$ are independent, where each $X_i$ has probability (mass) function $p_i$($x_i$) given as follows:

$p_i$($x_i$) = $\frac{e^{-\lambda}\lambda_i^{x_i}}{x_i!}$ (the parameter $\lambda_i$ differs in the distribution for each $X_i$ for $x_i$ = 0, 1, ...

What is the distribution of their sum $\Sigma_{i = 1}^{n}X_i$? Prove it using a moment generating function.

Also what is the approximate distribution of $\sum_{i=1}^{5}\frac{(X_i - \lambda_i)}{\lambda_i}$ if $X_1$, $X_2$, ... $X_5$ are very nearly normal?

Can somebody help me out with these two questions? I'm aware of the mgf quality $M_{\Sigma_{i = 1}^nX_i}(t) = M_{x_1}(t) * M_{x_2}(t) * ... * M_{x_n}(t)$ but I'm not sure how exactly I can use the resulting MGF to get the CDF or PDF.

$\endgroup$
  • $\begingroup$ Here's a hint: calculate the MGF of each individual rv, and then take their product. It's going to look very similar to what you originally calculated after you group all the constants together. And MGF's and distributions have a one-to-one correspondence. If you're still stuck, look up the Poisson distribution on wikipedia. $\endgroup$ – Alex R. Apr 24 '13 at 4:12
  • $\begingroup$ I calculated the MGFs and ended up with this $e^{\Sigma_{i=1}^{n}\lambda_i (e^t - 1)}$. So would the actual distribution be $p_i$($\Sigma_{i=1}^{n}x_i$) = $\frac{e^{-\Sigma_{i=1}^{n}\lambda_i}\Sigma_{i=1}^{n}\lambda_i^{x_i}}{x_i!}$? Is this correct? $\endgroup$ – tuba09 Apr 24 '13 at 4:23
  • $\begingroup$ Be careful. The MGF is of the sum. As in you are defining a new rv $S:=\sum_{i=1}^nX_i$ so that the product of MGF's gives you the MGF of $S$. Your MGF calculation is correct but your distribution should be in terms of $S$, not the individual $x_i$'s. $\endgroup$ – Alex R. Apr 24 '13 at 4:28
  • $\begingroup$ Alright I hear ya. I think i see what you're saying. Let me give this another shot. $\endgroup$ – tuba09 Apr 24 '13 at 4:37
  • $\begingroup$ Wait so would I first need to calculate the expected value using the MGF and then plug that in to the poisson distribution? How would I get the derivative of an MGF that has summation? $\endgroup$ – tuba09 Apr 24 '13 at 4:41
1
$\begingroup$

Hints: (1) A random variable $W$ has Poisson distribution with parameter $\lambda$ if and only if the distribution of $W$ has mgf $\exp\left(\lambda(e^t-1)\right)$.

(2) Since $X_i$ has mgf $\exp\left(\lambda_i(e^t-1)\right)$, by independence $\sum X_i$ has mgf the product of the individual mgf. This product simplifies to $\exp\left((\lambda_1+\lambda_2+\cdots +\lambda_n)(e^t-1)\right)$. , (3) Now look again at (1). The result we got through the mgf is that the sum of independent random variables with Poisson distribution has Poisson distribution.

For the second question, from the mgf, or otherwise, we find that the mean of $X_i$ is $\lambda_i$. Also from the mgf, we find that the expectation of $X_i^2$ is $\lambda+\lambda^2$, and therefore the variance of $X_i$ is $\lambda_i$.

Thus $\frac{X_i-\lambda_i}{\lambda_i}$ has mean $0$ and variance $\frac{1}{\lambda_i^2}\lambda_i=\frac{1}{\lambda_i}$.

The sum asked about is the sum of $5$ independent nearly normals. So the sum is nearly normal. From the known mean and variance of $\frac{X_i-\lambda_i}{\lambda_i}$ we conclude that $\sum_{i=1}^5 \frac{X_i-\lambda_i}{\lambda_i}$ is nearly normal, mean $0$, variance $\sum_{i=1}^5 \frac{1}{\lambda_i}$.

$\endgroup$
  • $\begingroup$ Hi Andre, I don't think there's any way for me to express how thankful I am. I've been pulling my hair over this question for so long but it makes perfect sense now. I just have one tiny unrelated question. How did you calculate the variance to be $\frac{1}{\lambda_i}$? I've never known how to calculate variance, I just know that if X ~ $N(\mu,\sigma^2)$ then $Z=\frac{X - \mu}{\sigma}$, Z ~ $N(0,1)$. How exactly are the 0 and 1 calculated, and how does this logic transfer over to what you just did there to calculate $\frac{1}{\lambda_i}$? $\endgroup$ – tuba09 Apr 24 '13 at 14:39
  • $\begingroup$ The variance of any $Y$ can be computed as $E(Y^2)-(E(Y))^2$. Then for the Poisson you can find $E(Y_i)$, $E(Y_i^2)$ from the mgf. There are $\lambda_i$, and $\lambda_i+\lambda_i^2$ respectively, so the variance of the Poisson $X_i$ is $\lambda_i$. Now for any random variable $Y$, and any constant $k$, the variance of $kY$ is $k^2$ times the variance of $Y$. So since the variance of $X_i-\lambda_i$ is $\lambda_i$, when we divide by $\lambda_i$ (mulriply by $\frac{1}{\lambda_i}$, the variance gets divided by $\lambda_i^2$. If you need detail about your question, please send message. $\endgroup$ – André Nicolas Apr 24 '13 at 14:56

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.