The result is not Ideal

[whecrane]

Thanks for your reply. When running the MTAG, I get an error message just like some others have, the mean chi2 statistic of triat 2 is less than 1.02, I used the --force. Fortunetly, there have the results. But which make me confused that the result is not as good as the original. The snp belong to p<5*10^-8 is less than the original. Why did I get this result? Does it work wrong because of the chi2<1.02? Thank you very much

[paturley]

Hello,

MTAG will run just fine if the mean chi2 is less than 1.02, but the results will tend to be unreliable. The first step in MTAG is to estimate the genetic correlation between the pairs of included traits and it uses that estimate in order to combine the summary statistics. If any of the sets of GWAS summary statistics are low powered, then the estimate of the genetic correlation will be very imprecise and MTAG will give unstable results. So you may end up with SNPs that were once significant that no longer are significant or SNPs that MTAG tells you are significant that are not actually associated with the trait of interest.

Hope that helps.

Best, Patrick

On Fri, Apr 3, 2020 at 2:18 AM whecrane [email protected] wrote:

Thanks for your reply. When running the MTAG, I get an error message just like some others have, the mean chi2 statistic of triat 2 is less than 1.02, I used the --force. Fortunetly, there have the results. But which make me confused that the result is not as good as the original. The snp belong to p<5*10^-8 is less than the original. Why did I get this result? Does it work wrong because of the chi2<1.02? Thank you very much

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[whecrane]

Hi Patrick, Thank you for your promptly reply. That's helpful. I will use some other summary statistics to have a try. By the way, if I just have Pvale and Beta and STAT (Coefficient t-statistic ), can I calculate the freq or the se column? Just like this: CHR Chromosome SNP SNP identifier BP Physical position (base-pair) A1 Tested allele (minor allele by default) TEST Code for the test (see below) NMISS Number of non-missing individuals included in analysis BETA/OR Regression coefficient (--linear) or odds ratio (--logistic) STAT Coefficient t-statistic P Asymptotic p-value for t-statistic

Best wishes, Crane

[paturley]

Hi Crane,

I'm sorry, I don't know if I understand your question. If you have the beta and t-statistics, you should be able to infer the standard error, but you wouldn't be able to infer the frequency. You could look up the allele frequency in some reference panel, however.

On Sat, Apr 4, 2020 at 11:37 AM whecrane [email protected] wrote:

Hi Patrick, Thank you for your promptly reply. That's helpful. I will use some other summary statistics to have a try. By the way, if I just have Pvale and Beta and STAT (Coefficient t-statistic ), can I calculate the freq or the se column? Just like this: CHR Chromosome SNP SNP identifier BP Physical position (base-pair) A1 Tested allele (minor allele by default) TEST Code for the test (see below) NMISS Number of non-missing individuals included in analysis BETA/OR Regression coefficient (--linear) or odds ratio (--logistic) STAT Coefficient t-statistic P Asymptotic p-value for t-statistic

Best wishes, Crane

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[whecrane]

Hi Patrick, Thank you for taking the time to answer questions , thank you. To your health.

Best wishes, Crane

[whecrane]

Hi Patrick, I used the ldsc to run the summary statistics, but the Mean chi^2 is bigger than which in the mtag, so it is 0.941 in mtag but 1.134 in ldsc . I don't know the difference is because N in ldsc is the sample size, and N in mtag is the size of snp? I am very confused about this and look forward to your answer.

Best wishes, Crane

[paturley]

Sorry I don't understand your question. Is it that MTAG and LDSC are reporting different mean chi2 statistics?

On Tue, Apr 7, 2020 at 12:01 AM whecrane [email protected] wrote:

Hi Patrick, I used the ldsc to run the summary statistics, but the Mean chi^2 is bigger than which in the mtag, so it is 0.941 in mtag but 1.134 in ldsc . I don't know the difference is because N in ldsc is the sample size, and N in mtag is the size of snp? I am very confused about this and look forward to your answer.

Best wishes, Crane

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[whecrane]

Yes, the mean chi2 is different between the mtag and the ldsc .The mtag is lower than ldsc

| | 王鹤 | | 邮箱：[email protected] | On 04/08/2020 10:25, paturley wrote:

Sorry I don't understand your question. Is it that MTAG and LDSC are reporting different mean chi2 statistics?

On Tue, Apr 7, 2020 at 12:01 AM whecrane [email protected] wrote:

Hi Patrick, I used the ldsc to run the summary statistics, but the Mean chi^2 is bigger than which in the mtag, so it is 0.941 in mtag but 1.134 in ldsc . I don't know the difference is because N in ldsc is the sample size, and N in mtag is the size of snp? I am very confused about this and look forward to your answer.

Best wishes, Crane

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[paturley]

I still don't understand where you are getting the numbers from. Can you attach the log files for LDSC and MTAG?

On Tue, Apr 7, 2020 at 10:35 PM whecrane [email protected] wrote:

Yes, the mean chi2 is different between the mtag and the ldsc .The mtag is lower than ldsc

| | 王鹤 | | 邮箱：[email protected] | On 04/08/2020 10:25, paturley wrote:

Sorry I don't understand your question. Is it that MTAG and LDSC are reporting different mean chi2 statistics?

On Tue, Apr 7, 2020 at 12:01 AM whecrane [email protected] wrote:

Hi Patrick, I used the ldsc to run the summary statistics, but the Mean chi^2 is bigger than which in the mtag, so it is 0.941 in mtag but 1.134 in ldsc . I don't know the difference is because N in ldsc is the sample size, and N in mtag is the size of snp? I am very confused about this and look forward to your answer.

Best wishes, Crane

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.

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[whecrane]

I'm sorry I didn't explain the problem, these are my log files: AD_test_ldsc.log PA_disease_mtag.log Parkinson_test_ldsc.log