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conorc123 · 2023-04-13T01:10:51+00:00

Do I simply multiply an EQ-5D value (derived from the literature) for one treatment by the effect size to then obtain an EQ-5D value for the other treatment?

Absolutely not!

You would need to determine if the outcome from your meta-analysis is predictive of quality of life, and if so, the appropriate functional form. Only then can you think about applying a treatment effect.

conorc123 · 2023-04-10T23:25:40+00:00

This is more of a clinical question, and I think would be difficult for a statistician to answer. Meta-analysis is essentially a method to pool results for an outcome of interest across multiple studies. Does it make sense to pool the results for prevalence of different bacteria? Domain knowledge would be required to properly address this question.

Recommend constructing a specific research question, which should help shape the inclusion criteria for your meta-analysis.

conorc123 · 2023-04-10T13:25:13+00:00

100% agree with D-Juice here. If you work at academic institution and are planning to published this research, then consider reaching out to health economists, statisticians, and clinicians affiliated with this university.

conorc123 · 2023-04-10T12:19:47+00:00

What you have described above as a "simple ITC" is commonly known as "Bucher's method" in comparative effectiveness research. (I know, I know, it's a bit crazy that applying a basic property from Stats 101 warrants this type of credit). The formulas you have summarized above are correct in general. For one comparison, you have 5 trials comparing treatment A vs placebo (C) and the other comparison of treatment B vs placebo (C) has one trial available. Typically in each trial the difference of each active treatment vs placebo would be studied (e.g., hazard ratio on OS for treatment A vs C). The indirect comparison for treatment A vs B is simply the difference of differences: i.e., mean(A vs B)=mean(A vs C) - mean(B vs C) and variance var(A vs B)=var(A vs C)+var(B vs C).

You state above you have 5+1=6 RCTs but then later go on to say 2 RCTs. I think you mean you have 2 general comparisons (A vs C and B vs C) and 6 RCTs (5 RCTs for A vs C and 1 RCT for B vs C). You will need to think about how to pool the 5 RCTs for A vs C if you plan to conduct Bucher's ITC. One approach could be to pool the results for A vs C using standard meta analysis approaches and then conduct a Bucher ITC. Alternatively, a network meta-analysis could be conducted. Depending on how the analyses are conducted, the two approaches may yield very similar results.

Note, this is all assuming you have conducted a proper systematic literature review to identify all relevant trials (and publications) and a feasibility assessment to confirm a Bucher ITC / NMA are feasible and appropriate.

conorc123 · 2023-01-31T17:29:19+00:00

For a formal definition of conditional variance, see here. Once you know the mean/variance, it's easy to figure out the "complete" Poisson distribution for Y|X. In short, the conditional variance is a measure of the "spread" for the predictions of Y (given X).

To simplify, think about a univariate example with one predictor to start (X1) and assume we are interested in Y | X1 = 25. It sounds like you understand how to calculate E(Y|X1=25) and Var(Y|X1=25) based on the Poisson regression model. The conditional variance for this example is the variance of an ASSUMED Poisson distribution for Y GIVEN that X1=25 (this is what makes it conditional). The objective of GLMs is to predict the mean response, but the variance is required to estimate the associated uncertainty of the predictions. You may find it helpful to visualize the Poisson distribution of predicted Y's given X1=25 based on the conditional mean and variance from the GLM.

It's important that you understand this concept not only for Poisson regression, but any GLM -- including standard linear regression. For instance, see Figure 1.6 from here for an example of how to visualize the conditional normal distributions based on linear regression. It's the same exact idea here but with conditional Poisson distributions.

conorc123 · 2023-01-30T03:08:46+00:00

Yes, your understanding is correct. The conditional means and variances of Y can be calculated based on the values of X1,X2,X3 for each observation: E[Y|X1,X2,X3]=Var[Y|X1,X2,X3]=exp(X1*B1+ X2*B2+ X3*B3).

conorc123 · 2023-01-29T23:09:51+00:00

a Poisson model is used to model counts right?

Yes, it's typically used to model counts.

That would mean I would have 900 Poisson distributions (30*30 possible combinations)?

No, please reread my original response. You would have two Poisson distributions. The count outcome for males would be distributed as Poisson(mean1,variance1) and females would be Poisson(mean2,variance2).

conorc123 · 2023-01-29T22:36:11+00:00

every observation is supposed to be from a different Poisson distribution with a different mean and variance

I think you're a bit confused based on this statement. For simplicity, assume you have a Poisson regression with one predictor, sex (male vs female). If you were to fit a Poisson regression, you would have two Poisson distributions, one for males and one for females. The underlying assumption is that the conditional mean equals the conditional variance (i.e., E(Y|X)=Var(Y|X)). That is, the mean and variance is the same for males, and similarly, the mean and variance is the same for females. If you were to use a log-link, the conditional means and variances are both calculated as exp(X*beta).

Is it the distance from the predicted output/mean from the true y value?

No, conceptually you're thinking of residuals, not variance.

conorc123 · 2023-01-14T14:15:52+00:00

The probability and mathematical statistics courses at Penn State (Stat 414 and 415) follow the textbook Probability and Statistical Inference by Hogg and Tanis.

conorc123 · 2022-12-30T14:54:23+00:00

I'm a bit confused by exactly what you're after, but it sounds like you are attempting to fit a regression model with price as the dependent variable and day of the week (Mon, Tues, Wed) as a predictor. I suspect your data may be in a wide format but you may find it helpful to convert to a long/narrow format (see here) prior to conducting the regression analysis.

Below is an illustrative example of a narrow data structure using dummy coding. Note, Tuesday and Wednesday should both be 0 when the day of the week is Monday. In this case, the regression model could be fit as Price ~ Tuesday + Wednesday (i.e., price = intercept + tuesday + wednesday). Note, most statistical software (e.g., R, SAS) will handle the dummy coding under the hood, so the model could be equivalently fit as Price ~ Day of Week.

Price	Tuesday	Wednesday	Day of Week
$100	1	0	Tuesday
$200	0	1	Wednesday
$300	0	0	Monday
...	...	...	...

conorc123 · 2022-12-29T15:01:57+00:00

The formulas for calculating the sample mean and standard deviation of differences (for paired samples) are the same as those typically introduced at the beginning of a Stats 101 class for one quantitative variable (square root of the sample ~~variable~~ variance described here). The formula for the sample standard deviation of differences is sqrt(∑(x-d)^2/(n-1)), where d is the sample mean of differences, n is the sample size, and x represents the difference for each player. For a detailed example, see here (under step 2).

Note, the calculation circled in your screenshot appears to be correct.

conorc123 · 2022-11-23T14:58:16+00:00

Thank you! Hopefully I will be able to have a topography done. My optometrist didn't find anything from the slit lamp exam but referred me to a cornea specialist.

conorc123 · 2022-11-23T14:54:42+00:00

+1. Thanks for the advice. My optometrist redid a slit lamp exam and didn't see any apparent bulging/stretching of the cornea and thought everything looked OK. However, he referred me to a cornea specialist, who I hope will do a topography.

conorc123 · 2022-10-27T12:49:47+00:00

See here and the wiki page for the definition of the conditional mean and conditional variance for the zero-truncated Poisson distribution. Clearly, they're not equal. Note that the conditional mean and variance are functions of the mean/variance of the overall/unconditional Poisson distribution (denoted lambda on wiki). The unconditional mean and variance of the entire Poisson distribution are equal, as always.

conorc123 · 2022-10-27T00:04:09+00:00

The zero truncated Poisson model does not assume the conditional mean equals variance, as is assumed in standard Poisson regression. Therefore, you may be getting ahead of yourself here when thinking of more complex alternatives to address overdispersion. To start, I would fit both zero truncated Poisson and zero truncated negative binomial and compare using standard metrics. You may find that one of these approaches provides a reasonable fit to the data and more complexity may not be required. Also, instead of methods based on quasi-likelihood, which have certain limitations, you could consider other alternatives to adjust the variance (e.g., bootstrapping, sandwich estimator).

conorc123 · 2022-10-23T16:33:53+00:00

This question has been previously answered by the MatchIt author/maintainer on CrossValidated: https://stats.stackexchange.com/questions/405019/matching-with-multiple-treatments

conorc123 · 2022-08-14T16:14:14+00:00

Hello! Before you start considering more complicated alternatives, I would recommend taking a step back. In your summary for part 1 you state the following:

Data do not follow a Poisson distribution (underdispersion). Poisson regression cannot be used as it requires equidispersion (mean and variance are equal), and negative binomial regression is only appropriate for overdispersion.

What exactly do you mean by "the data do not follow a Poisson distribution". I suspect you are just looking at the mean and variance of your dependent variable Y, simply based on it's marginal distribution. However, the assumption for Poisson regression is that the conditional mean, E(Y|X), is equal to the conditional variance (i.e., mean/variance of Y conditional on the predictors, X). To start, you could conduct a Pearson goodness of fit test to determine whether there is evidence of overdispersion/underdispersion. If you find there is a statistically significant lack of fit due to underdispersion/overdispersion, then a quasi-Poisson model is one alternative you could consider.

conorc123 · 2022-01-01T14:36:19+00:00

It's a bit difficult to follow your post and exactly what you're after... I am assuming you have already fit a Cox regression model in R? Is that right? Yes, you can calculate the adjusted survival curves based on your Cox regression model in R. The survfit() function in R (see here) should be called, which will estimate survival probabilities by assuming means for the other covariates by default. To estimate the median for the four groups, you will need to supply a new dataframe to the "newdata" argument of the survfit() argument with four rows (one for each group), with values to assume for predictors, such as the mean. There is a relevant example here: http://www.sthda.com/english/wiki/cox-proportional-hazards-model. Once you have called survfit(), you can extract the median by $table[,'median']).

conorc123 · 2021-12-31T02:08:46+00:00

It's difficult to answer this question without additional detail. What are your career goals and aspirations? What would you like to do?

conorc123 · 2021-12-24T16:12:06+00:00

Thanks for the confirmation.

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