Solve a strange inequality

Homework Statement

Solving an exercise I found myself with this problem: the solution $c$ needs to verify both $\sum_{k=1}^c n\lambda^k\frac{e^{n\lambda}}{k!}\leq \alpha$ and $1-\sum_{k=1}^{c+1} n\lambda^k\frac{e^{n\lambda}}{k!}\geq \alpha$.

Can an equation like this be solved for c?

Homework Equations

The Attempt at a Solution

An inversion of the inequality didn't work because it vanishes $\alpha$, I don't think there is an explicit form for the solution, but I've failed in finding an argument to explain why.

 

[mfb]

Assuming all parameters are positive, both sums increase with increasing c, so the inequalities will be valid from c=0 (or c=1) up to some maximal c. If you are just interested in finding any solution, test the smallest c allowed.

Apart from constant prefactors, your sum is $\displaystyle \sum_{k=1}^{c} \frac{\lambda^k}{k}$, I'm not aware of closed forms for that (although some special values of $\lambda$ might have them). Is it really $k$ in the denominator, not $k!$ ?

 

Assuming all parameters are positive, both sums increase with increasing c, so the inequalities will be valid from c=0 (or c=1) up to some maximal c. If you are just interested in finding any solution, test the smallest c allowed.

Apart from constant prefactors, your sum is $\displaystyle \sum_{k=1}^{c} \frac{\lambda^k}{k}$, I'm not aware of closed forms for that (although some special values of $\lambda$ might have them). Is it really $k$ in the denominator, not $k!$ ?

It is k!. I mistyped that.

There was another error, it is
$1-\sum_{k=1}^{c} n^k\lambda^k\frac{e^{n\lambda}}{k!}\geq \alpha$, NOT $1-\sum_{k=1}^{c+1} n\lambda^k\frac{e^{n\lambda}}{k!}\geq \alpha$.

and

$\sum_{k=1}^{c} n^k\lambda^k\frac{e^{n\lambda}}{k!}\leq \alpha$ instead of $\sum_{k=1}^{c} n\lambda^k\frac{e^{n\lambda}}{k!}\leq \alpha$

 

[mfb]

Okay, so we have the standard Poisson distribution with $n\lambda$ expectation value. I'm not aware of closed formulas for the cumulative distribution function.

As the two sums are now the same, the problem got easier: Check which inequality you have to consider for $\alpha$ smaller or larger than 1/2,

 

Okay, so we have the standard Poisson distribution with $n\lambda$ expectation value. I'm not aware of closed formulas for the cumulative distribution function.

As the two sums are now the same, the problem got easier: Check which inequality you have to consider for $\alpha$ smaller or larger than 1/2,

I don't understand the last part. I don't have the values $n,\lambda,c$ which I think I need to calculate the sum, and I need to calculate the sum in order too know which inequality should I use if $\alpha$ smaller or larger than 1/2.

 

[fresh_42]

Without all the constants in the sum, it is just $\sum_{k=1}^c \frac{\lambda^k}{k!} \leq \ldots$ which is just the exponential function $- 1$. There are probably more approximations $r$ for $e^x = \sum_{k=0}^N \frac{x^k}{k!} + r(x;N) \text{ with } r(x;N) \leq \ldots$ than of any other function.

 

[Ray Vickson]

It is k!. I mistyped that.

There was another error, it is
$1-\sum_{k=1}^{c} n^k\lambda^k\frac{e^{n\lambda}}{k!}\geq \alpha$, NOT $1-\sum_{k=1}^{c+1} n\lambda^k\frac{e^{n\lambda}}{k!}\geq \alpha$.

and

$\sum_{k=1}^{c} n^k\lambda^k\frac{e^{n\lambda}}{k!}\leq \alpha$ instead of $\sum_{k=1}^{c} n\lambda^k\frac{e^{n\lambda}}{k!}\leq \alpha$

Do you really mean to have $e^{n \lambda}$ rather than $e^{- n \lambda}$? I ask, because if $p_k(\mu) = \mu^k e^{-\mu}/k!$ is the Poisson probability mass function for mean $\mu$ your sum has the form $e^{2 n \lambda} \sum_{k=1}^c p_k(n \lambda)$. If you had $e^{-n \lambda}$ instead, your sum would be $\sum_{k=1}^c p_k(n \lambda)$, which is almost the cumulative distribution function; it merely lacks the $k=0$ term.

 

Do you really mean to have $e^{n \lambda}$ rather than $e^{- n \lambda}$? I ask, because if $p_k(\mu) = \mu^k e^{-\mu}/k!$ is the Poisson probability mass function for mean $\mu$ your sum has the form $e^{2 n \lambda} \sum_{k=1}^c p_k(n \lambda)$. If you had $e^{-n \lambda}$ instead, your sum would be $\sum_{k=1}^c p_k(n \lambda)$, which is almost the cumulative distribution function; it merely lacks the $k=0$ term.

You're right. It is $e^{-n\lambda}$.

It's supposed to be the cumulative function. This is part of a larger problem, I didn't post the whole thing because I had some parts worked out. I don't think I can edit the initial post anymore, but I will post the original problem to give some context (which I probably should have done before).

The idea of the problem consists in having $X_1,\dots, X_n$ simple random sample of $X\sim \operatorname{Poisson}(\lambda)$ and the goal is the optimal critical region $\alpha$ for $H_0:\lambda =\lambda_0$ against $H_1:\lambda=\lambda_1$.

I applied Neyman-Pearson theorem and followed $\frac{\lambda_0}{\lambda_1} = \frac{(\lambda_0^{\sum x_i} e^{-\lambda_0}) / (\prod_{i=1}^n x_i!)}{(\lambda_1^{\sum x_i}e^{-\lambda_1})/(\prod_{i=1}^n x_i!)} = \frac{\lambda_0^{\sum x_i}e^{\lambda_1}}{\lambda_1^{\sum x_i}e^{\lambda_0}}\leq k$

Taking logarithms

$\ln \left( \lambda_0^{\sum x_i}e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}e^{\lambda_0} \right)\leq \ln(k)\\ \ln \left( \lambda_0^{\sum x_i}\right)+\ln\left(e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}\right)-\left(e^{\lambda_0} \right)\leq \ln(k)\\ \left(\sum x_i\right)\ln (\lambda_0)+\lambda_1 -\left( \sum x_i \right)\ln \left( \lambda_1\right)-\lambda_0\leq \ln(k)\\ \left(\sum x_i\right)\left(\ln (\lambda_0) -\ln \left( \lambda_1\right)\right)\leq \ln(k)+(\lambda_0-\lambda_1)\\ \sum x_i\leq \frac{\ln(k)+(\lambda_0-\lambda_1)}{\ln (\lambda_0) -\ln \left( \lambda_1\right)}$

Because of that I have to find $c$ such that
$P\left( \sum_{i=1}^n X_i \leq c |, \lambda = \lambda_0 \right) \leq \alpha$

and

$P\left( \sum_{i=1}^{c+1} X_i \le c \,\middle|\, \lambda = \lambda_0 \right) \geq \alpha$


Which reduced to the pair of equations you see above.

 

[Ray Vickson]

You are right, it is $e^{-n\lambda}$

In that case, you can express the sum in terms of known functions:
$$\sum_{k=0}^c \frac{(n \lambda)^k e^{-n \lambda}}{k!} = \frac{\Gamma(c+1,n \lambda)}{\Gamma(c+1)}, $$
where $\Gamma(p,q)$ is the incomplete Gamma function:
$$\Gamma(p,q) = \int_q^{\infty} t^{p-1} e^{-t} \, dt .$$
Whether this is of any use at all is another issue.

 

[Ray Vickson]

You're right. It is $e^{-n\lambda}$.

It's supposed to be the cumulative function. This is part of a larger problem, I didn't post the whole thing because I had some parts worked out. I don't think I can edit the initial post anymore, but I will post the original problem to give some context (which I probably should have done before).

The idea of the problem consists in having $X_1,\dots, X_n$ simple random sample of $X\sim \operatorname{Poisson}(\lambda)$ and the goal is the optimal critical region $\alpha$ for $H_0:\lambda =\lambda_0$ against $H_1:\lambda=\lambda_1$.

I applied Neyman-Pearson theorem and followed $\frac{\lambda_0}{\lambda_1} = \frac{(\lambda_0^{\sum x_i} e^{-\lambda_0}) / (\prod_{i=1}^n x_i!)}{(\lambda_1^{\sum x_i}e^{-\lambda_1})/(\prod_{i=1}^n x_i!)} = \frac{\lambda_0^{\sum x_i}e^{\lambda_1}}{\lambda_1^{\sum x_i}e^{\lambda_0}}\leq k$

Taking logarithms

$\ln \left( \lambda_0^{\sum x_i}e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}e^{\lambda_0} \right)\leq \ln(k)\\ \ln \left( \lambda_0^{\sum x_i}\right)+\ln\left(e^{\lambda_1} \right)-\ln \left( \lambda_1^{\sum x_i}\right)-\left(e^{\lambda_0} \right)\leq \ln(k)\\ \left(\sum x_i\right)\ln (\lambda_0)+\lambda_1 -\left( \sum x_i \right)\ln \left( \lambda_1\right)-\lambda_0\leq \ln(k)\\ \left(\sum x_i\right)\left(\ln (\lambda_0) -\ln \left( \lambda_1\right)\right)\leq \ln(k)+(\lambda_0-\lambda_1)\\ \sum x_i\leq \frac{\ln(k)+(\lambda_0-\lambda_1)}{\ln (\lambda_0) -\ln \left( \lambda_1\right)}$

Because of that I have to find $c$ such that
$P\left( \sum_{i=1}^n X_i \leq c |, \lambda = \lambda_0 \right) \leq \alpha$

and

$P\left( \sum_{i=1}^{c+1} X_i \le c \,\middle|\, \lambda = \lambda_0 \right) \geq \alpha$


Which reduced to the pair of equations you see above.

I really don't "get" these two equations. Suppose, eg., that $\lambda_0 < \lambda_1$. Say we accept the null hypothesis if $\sum X_i \leq c$ and reject it if $\sum X_i > c$. (Since the random variables are discrete we need to distinguish $< c$ from $\leq c$, and I have more-or-less arbitrarily chosen the $\leq c$ version for the acceptance region.) Thus, we accept $H_0$ if $\sum_{I=1}^n X_i \leq c$ and reject it otherwise. So, the probability of a type-I error is $P(I) = P(\sum_{I=1}^n X_i > c| \lambda = \lambda_0)$, and we want this to be $\leq \alpha.$ For a given $n$ that tells you the acceptable values of $c$.

The probability of a type-II error is $P(II) = P(\sum_{I=1}^n X_i \leq c | \lambda = \lambda_1)$, and (presumably) we want this to be $\leq \beta$ for some specified value $\beta.$ Your presentation does mention any $\beta$, but of course you could take $\beta = \alpha.$ However, if you are doing that you need to mention it!

The two conditions taken together give two inequalities involving $(n , c)$, which we can probably solve (numerically) using a computer algebra package such as Maple or Mathematica.

 

[StoneTemplePython]

You're right. It is $e^{-n\lambda}$.
I will post the original problem to give some context (which I probably should have done before).

You definitely should have.

You're right. It is $e^{-n\lambda}$.
The idea of the problem consists in having $X_1,\dots, X_n$ simple random sample of $X\sim \operatorname{Poisson}(\lambda)$ and the goal is the optimal critical region $\alpha$ for $H_0:\lambda =\lambda_0$ against $H_1:\lambda=\lambda_1$.

Suppose for a minute that $X$ is instead a normal random variable with variance of 1. Can you solve this? How would you do it?

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I am getting concerned that you are getting into the weeds of classical stats with a firm grounding in probability. Classical stats is already notorious for being presented in a cookbook of recipes fashion, with ties to probability, making it confusing for people. If the ties aren't strong, it gets worse.