多項ロジスティック回帰におけるexp(B)の解釈


16

これは多少初心者の質問ですが、多項ロジスティック回帰モデルで6.012のexp(B)結果をどのように解釈するのでしょうか?

1)6.012-1.0 = 5.012 = 5012%のリスク増加ですか?

または

2)6.012 /(1 + 6.012)= 0.857 = 85.7%のリスク増加?

両方の選択肢が間違っている場合、誰かが正しい方法を言及してもらえますか?

私はインターネットで多くのリソースを検索しましたが、これらの2つの選択肢にたどり着きましたが、どちらが正しいか完全にはわかりません。

回答:


35

そこに到達するにはしばらく時間がかかりますが、要約すると、Bに対応する変数の1単位の変更は、結果の相対リスク(基本結果と比較)に6.012を掛けます。

これを相対リスクの「5012%」増加として表現することもできますが、実際には多項ロジスティックモデルが強く推奨しているのに、変更を相加的に考える必要があることを示唆しているため、混乱を招く可能性があり、誤解を招く可能性があります乗法的に考える。変数の変更は、問題の結果だけでなく、すべての結果の予測確率を同時に変更するため、修飾子「相対」が不可欠です。そのため、確率を比較する必要があります(差はなく比率を使用)。

この応答の残りの部分では、これらのステートメントを正しく解釈するために必要な用語と直感を開発します。

バックグラウンド

多項のケースに移る前に、通常のロジスティック回帰から始めましょう。

従属(バイナリ)変数Yおよび独立変数Xi場合、モデルは

Pr[Y=1]=exp(β1X1++βmXm)1+exp(β1X1++βmXm);

同様に、0 Pr [ Y = 1と仮定0Pr[Y=1]1と、

log(ρ(X1,,Xm))=logPr[Y=1]Pr[Y=0]=β1X1++βmXm.

(これは単純に定義しρ、これはX iの関数としてのオッズです。)Xi

一般に、インデックスの損失なしXiその結果Xm可変であり、βm当該「B」である(その結果、exp(βm)=6.012)。値は固定Xi,1i<m、及び変化Xm少量δ収率

log(ρ(,Xm+δ))log(ρ(,Xm))=βmδ.

したがって、βm に対して対数オッズに限界変化であります Xm

回復するにはexp(βm)、明らかに我々は設定しなければなりませんδ=1と左側の累乗します:

exp(βm)=exp(βm×1)=exp(log(ρ(,Xm+1))log(ρ(,Xm)))=ρ(,Xm+1)ρ(,Xm).

この展示としてオッズ比で1単位の増加のためにXのM。これが何を意味するのか直観を開発するには、開始オッズの範囲のいくつかの値を表にし、パターンを際立たせるために大きく丸めます:exp(βm)Xm

Starting odds  Ending odds  Starting Pr[Y=1]  Ending Pr[Y=1]
0.0001         0.0006       0.0001            0.0006
0.001          0.006        0.001             0.006
0.01           0.06         0.01              0.057
0.1            0.6          0.091             0.38
1.             6.           0.5               0.9
10.            60.          0.91              1.
100.           600.         0.99              1.

ために本当に小さなオッズ、対応するように本当に小さい確率の1つの単位の増加の効果する乗算 6.012程度によってオッズ又は確率。乗法係数、オッズ(および確率)が大きくなるにつれて減少し、オッズが10を超えると実質的に消滅します(確率は0.9を超えます)。Xm

Ratio change in probability

相加的な変化として、0.0001と0.0006の確率に大きな差はなく(0.05%のみ)、0.99と1の差はほとんどありません(1%のみ)。オッズが1 / √の場合、最大の加算効果が発生します、ここで、29%から71%の確率の変更:+ 42%の変化。1/6.0120.408

Additive change in probability

我々はオッズ比として「リスク」を表現する場合、こと、次いで、参照オッズ比は等しく- =「B」は、単純な解釈を有するβ Mの単位増加のためにβmβm我々は、リスクを発現する場合-しかしを確率の変化など、他の方法では、解釈には開始確率を指定する注意が必要です。Xm

多項ロジスティック回帰

(これは後の編集として追加されました。)

対数オッズを使用してチャンスを表現することの価値を認識したら、多項のケースに移りましょう。今従属変数のいずれかと等しくすることができ、K 2つのによってインデックス付けカテゴリ、iは= 1 2 ... kは相対的確率は、それがカテゴリであることを私はありますYk2i=1,2,,ki

Pr[Yi]exp(β1(i)X1++βm(i)Xm)

パラメータを決定し、Pr [ Y = category  i ]にY iを書き込む。省略形として、右辺の式をp iX β として書くか、Xβがコンテキストから明らかな場合は、単にp iとします。正規化により、これらすべての相対確率の合計が1になります。βj(i)YiPr[Y=category i]pi(X,β)Xβpi

Pr[Yi]=pi(X,β)p1(X,β)++pm(X,β).

(パラメータにはあいまいさがあります。パラメータが多すぎます。従来は、比較のために「ベース」カテゴリを選択し、その係数をすべて強制的にゼロにします。ただし、これは、されていない係数を解釈するのに必要な対称性を維持するために- 。、あるカテゴリの中から任意の人工の区別を避けるために-私たちが持っていない限り、LETはそのような制約を強制していません)。

One way to interpret this model is to ask for the marginal rate of change of the log odds for any category (say category i) with respect to any one of the independent variables (say Xj). That is, when we change Xj by a little bit, that induces a change in the log odds of Yi. We are interested in the constant of proportionality relating these two changes. The Chain Rule of Calculus, together with a little algebra, tells us this rate of change is

 log odds(Yi) Xj=βj(i)βj(1)p1++βj(i1)pi1+βj(i+1)pi+1++βj(k)pkp1++pi1+pi+1++pk.

This has a relatively simple interpretation as the coefficient βj(i) of Xj in the formula for the chance that Y is in category i minus an "adjustment." The adjustment is the probability-weighted average of the coefficients of Xj in all the other categories. The weights are computed using probabilities associated with the current values of the independent variables X. Thus, the marginal change in logs is not necessarily constant: it depends on the probabilities of all the other categories, not just the probability of the category in question (category i).

k=2i=2βj(2)βj(1)iβj(2), because we force βj(1)=0. Thus the new interpretation generalizes the old.

To interpret βj(i) directly, then, we will isolate it on one side of the preceding formula, leading to:

The coefficient of Xj for category i equals the marginal change in the log odds of category i with respect to the variable Xj, plus the probability-weighted average of the coefficients of all the other Xj for category i.

Another interpretation, albeit a little less direct, is afforded by (temporarily) setting category i as the base case, thereby making βj(i)=0 for all the independent variables Xj:

The marginal rate of change in the log odds of the base case for variable Xj is the negative of the probability-weighted average of its coefficients for all the other cases.

Actually using these interpretations typically requires extracting the betas and the probabilities from software output and performing the calculations as shown.

Finally, for the exponentiated coefficients, note that the ratio of probabilities among two outcomes (sometimes called the "relative risk" of i compared to i) is

YiYi=pi(X,β)pi(X,β).

Let's increase Xj by one unit to Xj+1. This multiplies pi by exp(βj(i)) and pi by exp(βj(i)), whence the relative risk is multiplied by exp(βj(i))/exp(βj(i)) = exp(βj(i)βj(i)). Taking category i to be the base case reduces this to exp(βj(i)), leading us to say,

The exponentiated coefficient exp(βj(i)) is the amount by which the relative risk Pr[Y=category i]/Pr[Y=base category] is multiplied when variable Xj is increased by one unit.


1
Great explanations, but the OP explicitly asked for the multinomial model. I may be reading more into the question than the OP intended, and the explanation for the binary case may be adequate, but I would love to see this answer cover the general multinomial case too. Even though the parametrization is similar, the "log-odds" are in general with respect to an (arbitrary) reference category, and they are not really log-odds, and a unit change in Xi results in a combined change of these "log-odds", and an increasing "log-odds" does not imply and increasing probability.
NRH

@NRH That's an excellent point. I had somehow read "multivariate" instead of "multinomial." If I get a chance to return to this I will try to flesh out those details. Fortunately the same mode of analysis is effective in finding the correct interpretation.
whuber

@NRH Done. I welcome your suggestions (or anyone else's) about how to make the interpretation clearer, or for alternative interpretations.
whuber

1
thanks for writing this down. The complete answer is a very good reference.
NRH

1

Try considering this bit of explanation in addition to what @whuber has already written so well. If exp(B) = 6, then the odds ratio associated with an increase of 1 on the predictor in question is 6. In a multinomial context, by "odds ratio" we mean the ratio of these two quantities: a) the odds (not probability, but rather p/[1-p]) of a case taking the value of the dependent variable indicated in the output table in question, and b) the odds of a case taking the reference value of the dependent variable.

You seem to be looking to quantify the probability--rather than odds-- of a case being in one or the other category. To do this you would need to know what probabilities the case "started with" -- i.e., before we assumed the increase of 1 on the predictor in question. Ratios of probabilities will vary case by case, while the ratio of odds connected with an increase of 1 on the predictor stays the same.


"If exp(B) = 6, then the odds ratio associated with an increase of 1 on the predictor in question is 6", if I read @whuber's answer correctly it says that the odds ratio will be multiplied by 6 with an increase of 1 on the predictor. That is, the new odds ratio will not be 6. Or am I intepreting things incorrectly?
rbm

Where you say "the new odds ratio will not be 6" I would say "the new odds will not be 6...but the ratio of the new to the old odds will be 6."
rolando2

Yes, I agree with that! But I just thought that "the odds ratio associated with an increase of 1 on the predictor in question is 6" does not really say that. But maybe I am just misinterpreting it then. Thanks for the clarification!
rbm

1

I was also looking for the same answer, but the once above were not satisfying for me. It seemed to complex for what it really is. So I will give my interpretation, please correct me if I am wrong.

Do however read to the end, since it is important.

First of all the values B and Exp(B) are the once you are looking for. If the B is negative your Exp(B) will be lower than one, which means odds decrease. If higher the Exp(B) will be higher than 1, meaning odds increase. Since you are multiplying by the factor Exp(B).

Unfortunately you are not there yet. Because in a multinominal regression your dependent variable has multiple categories, let's call these categories D1, D2 and D3. Of which your last is the reference category. And let's assume your first independent variable is sex (males vs females).

Let's say the output for D1 -> males is exp(B)= 1.21, this means for males the odds increase by a factor 1.21 for being in the category D1 rather than D3 (reference category) compared to females (reference category).

So you are always comparing against your reference category of the dependent but also independent variables. This is not true if you have a covariate variable. In that case it would mean; a one unit increase in X increases the odds by a factor of 1.21 of being in category D1 rather than D3.

For those with an ordinal dependent variable:

If you have an ordinal dependent variable and did not do an ordinal regression because of the assumption of proportional odds for instance. Keep in mind your highest category is the reference category. Your result as above are valid to report. But keep in mind that an increase in odds than in fact means an increase in odds of being in the lower category rather than the higher! But that's only if you have an ordinal dependant variable.

If you want to know the increase in percentage, well take a fictive odds-number, let's say 100 and multiply it by 1.21 which is 121? Compared to 100 how much did it change percentage wise?


0

Say that exp(b) in an mlogit is 1.04. if you multiply a number by 1.04, then it increases by 4%. That is the relative risk of being in category a instead of b. I suspect that part of the confusion here might have to do with by 4% (multiplicative meaning) and by 4 percent points (additive meaning). The % interpretation is correct if we talk about a percentage change not percentage point change. (The latter would not make sense anyhow as relative risks aren't expressed in terms of percentages.)

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