# Thread: What is the distribution of ratios?

1. ## What is the distribution of ratios?

Hi,

would appreciate the help.

I am wondering if we have $X_1, X_2,X_3,X_4$ with $Gamma(a, \theta_1)$ and another distribution with $Y_1, Y_2 Gamma(b,\theta_2)$ what would be the distribution of $Y/X$ (Y and X are sample means) ?

I know it would be Gamma, but not sure about the new alpha and Beta?

Thanks

2. Originally Posted by hitman
Hi,

would appreciate the help.

I am wondering if we have $X_1, X_2,X_3,X_4$ with $Gamma(a, \theta_1)$ and another distribution with $Y_1, Y_2 Gamma(b,\theta_2)$ what would be the distribution of $Y/X$ (Y and X are sample means) ?

I know it would be Gamma, but not sure about the new alpha and Beta?

Thanks
First get the distribution of $\overline{Y}$ and $\overline{X}$. Are the $X_i$ independent and the $Y_i$ independent (using the moment generating function might then be one way of doing this).

Then get the cdf of $U = \frac{\overline{Y}}{\overline{X}}$.

Get the pdf from the cdf.

3. Well, the sum if indep gamma's with the same beta is a gamma.
The ratio of this random variable is similar to an F density.
Thats a ratio of $\chi^2$ random variables divided by it's degrees of freedom.

4. Originally Posted by matheagle
Well, the sum if indep gamma's with the same beta is a gamma.
The ratio of this random variable is similar to an F density.
Thats a ratio of $\chi^2$ random variables divided by it's degrees of freedom.
Ah,, I am confused $U_1/d_1/U_2/d_2$ U must have $\chi^2$ distribution. But I don't understand how do X bar and Y bar have $\chi^2$ distribution?

Thanks

5. Originally Posted by hitman
Ah,, I am confused $U_1/d_1/U_2/d_2$ U must have $\chi^2$ distribution. But I don't understand how do X bar and Y bar have $\chi^2$ distribution?

Thanks
They don't. I think you have misunderstood matheagles's reply.

1. $\overline{X} = \frac{X_1 + X_2 + X_3 + X_4}{4}$.

2. $m_{X_i}(t) = \frac{1}{(1 - \theta t)^a}$ therefore $m_{\frac{X_i}{4}}(t) = \frac{1}{\left(1 - \left(\frac{\theta}{4}\right) t\right)^a}$.

3. $m_{\overline{X}}(t) = m_{\frac{X_1}{4}}(t) \cdot m_{\frac{X_2}{4}}(t) \cdot m_{\frac{X_3}{4}}(t) \cdot m_{\frac{X_4}{4}}(t)$

(assuming independence)

$= \left[ \frac{1}{\left(1 - \left(\frac{\theta}{4}\right) t\right)^a}\right]^4$ $= \frac{1}{ \left(1 - \left( \frac{\theta}{4}\right) t\right)^{4a}}$

which is recognised as the mgf of another gamma distribution.

6. I see,, I think I got the idea, I am going to play with it.

Another question, about F-distribution, lets say $w \sim F(4,5)$ now we want to find a and b such that $P(W> a) = 0.05 = P(W

This question is not clear for me. $p(w>a) = 0.05$ is easy. you calculate the degree of freedom, look into the table for 3,4 and 0.05 and you find a. However, I am not sure about the next part. is it like:

$p(wb) \rightarrow p(w>b) = 0.95$

and probably to find 0.95 I have to use statistic software

7. Originally Posted by hitman
I see,, I think I got the idea, I am going to play with it.

Another question, about F-distribution, lets say $w \sim F(4,5)$ now we want to find a and b such that $P(W> a) = 0.05 = P(W

This question is not clear for me. $p(w>a) = 0.05$ is easy. you calculate the degree of freedom, look into the table for 3,4 and 0.05 and you find a. However, I am not sure about the next part. is it like:

$p(wb) \rightarrow p(w>b) = 0.95$

and probably to find 0.95 I have to use statistic software Mr F says: Yes, you do.
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